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4.9: Strategies for Solving Applications and Equations

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• Page ID 113697

Learning Objectives

By the end of this section, you will be able to:

• Use a problem solving strategy for word problems
• Solve number word problems
• Solve percent applications
• Solve simple interest applications

Before you get started, take this readiness quiz.

• Translate “six less than twice x ” into an algebraic expression. If you missed this problem, review [link] .
• Convert 4.5% to a decimal. If you missed this problem, review [link] .
• Convert 0.6 to a percent. If you missed this problem, review [link] .

Have you ever had any negative experiences in the past with word problems? When we feel we have no control, and continue repeating negative thoughts, we set up barriers to success. Realize that your negative experiences with word problems are in your past. To move forward you need to calm your fears and change your negative feelings.

Start with a fresh slate and begin to think positive thoughts. Repeating some of the following statements may be helpful to turn your thoughts positive. Thinking positive thoughts is a first step towards success.

• I think I can! I think I can!
• While word problems were hard in the past, I think I can try them now.
• I am better prepared now—I think I will begin to understand word problems.
• I am able to solve equations because I practiced many problems and I got help when I needed it—I can try that with word problems.
• It may take time, but I can begin to solve word problems.
• You are now well prepared and you are ready to succeed. If you take control and believe you can be successful, you will be able to master word problems.

Use a Problem Solving Strategy for Word Problems

Now that we can solve equations, we are ready to apply our new skills to word problems. We will develop a strategy we can use to solve any word problem successfully.

EXAMPLE \(\PageIndex{1}\)

Normal yearly snowfall at the local ski resort is 12 inches more than twice the amount it received last season. The normal yearly snowfall is 62 inches. What was the snowfall last season at the ski resort?

EXAMPLE \(\PageIndex{2}\)

Guillermo bought textbooks and notebooks at the bookstore. The number of textbooks was three more than twice the number of notebooks. He bought seven textbooks. How many notebooks did he buy?

He bought two notebooks

EXAMPLE \(\PageIndex{3}\)

Gerry worked Sudoku puzzles and crossword puzzles this week. The number of Sudoku puzzles he completed is eight more than twice the number of crossword puzzles. He completed 22 Sudoku puzzles. How many crossword puzzles did he do?

He did seven crosswords puzzles

We summarize an effective strategy for problem solving.

PROBLEM SOLVING STRATEGY FOR WORD PROBLEMS

• Read the problem. Make sure all the words and ideas are understood.
• Identify what you are looking for.
• Name what you are looking for. Choose a variable to represent that quantity.
• Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
• Solve the equation using proper algebra techniques.
• Check the answer in the problem to make sure it makes sense.
• Answer the question with a complete sentence.

Solve Number Word Problems

We will now apply the problem solving strategy to “number word problems.” Number word problems give some clues about one or more numbers and we use these clues to write an equation. Number word problems provide good practice for using the Problem Solving Strategy.

EXAMPLE \(\PageIndex{4}\)

The sum of seven times a number and eight is thirty-six. Find the number.

Did you notice that we left out some of the steps as we solved this equation? If you’re not yet ready to leave out these steps, write down as many as you need.

EXAMPLE \(\PageIndex{5}\)

The sum of four times a number and two is fourteen. Find the number.

EXAMPLE \(\PageIndex{6}\)

The sum of three times a number and seven is twenty-five. Find the number.

Some number word problems ask us to find two or more numbers. It may be tempting to name them all with different variables, but so far, we have only solved equations with one variable. In order to avoid using more than one variable, we will define the numbers in terms of the same variable. Be sure to read the problem carefully to discover how all the numbers relate to each other.

EXAMPLE \(\PageIndex{7}\)

The sum of two numbers is negative fifteen. One number is nine less than the other. Find the numbers.

EXAMPLE \(\PageIndex{8}\)

The sum of two numbers is negative twenty-three. One number is seven less than the other. Find the numbers.

\(−15,−8\)

EXAMPLE \(\PageIndex{9}\)

The sum of two numbers is negative eighteen. One number is forty more than the other. Find the numbers.

\(−29,11\)

Consecutive Integers (optional)

Some number problems involve consecutive integers . Consecutive integers are integers that immediately follow each other. Examples of consecutive integers are:

\[\begin{array}{rrrr} 1, & 2, & 3, & 4 \\ −10, & −9, & −8, & −7\\ 150, & 151, & 152, & 153 \end{array}\]

Notice that each number is one more than the number preceding it. Therefore, if we define the first integer as n , the next consecutive integer is \(n+1\). The one after that is one more than \(n+1\), so it is \(n+1+1\), which is \(n+2\).

\[\begin{array}{ll} n & 1^{\text{st}} \text{integer} \\ n+1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; & 2^{\text{nd}}\text{consecutive integer} \\ n+2 & 3^{\text{rd}}\text{consecutive integer} \;\;\;\;\;\;\;\; \text{etc.} \end{array}\]

We will use this notation to represent consecutive integers in the next example.

EXAMPLE \(\PageIndex{10}\)

Find three consecutive integers whose sum is \(−54\).

EXAMPLE \(\PageIndex{11}\)

Find three consecutive integers whose sum is \(−96\).

\(−33,−32,−31\)

EXAMPLE \(\PageIndex{12}\)

Find three consecutive integers whose sum is \(−36\).

\(−13,−12,−11\)

Now that we have worked with consecutive integers, we will expand our work to include consecutive even integers and consecutive odd integers . Consecutive even integers are even integers that immediately follow one another. Examples of consecutive even integers are:

\[24, 26, 28\]

\[−12,−10,−8\]

Notice each integer is two more than the number preceding it. If we call the first one n , then the next one is \(n+2\). The one after that would be \(n+2+2\) or \(n+4\).

Consecutive odd integers are odd integers that immediately follow one another. Consider the consecutive odd integers 63, 65, and 67.

\[63, 65, 67\]

\[n,n+2,n+4\]

Does it seem strange to have to add two (an even number) to get the next odd number? Do we get an odd number or an even number when we add 2 to 3? to 11? to 47?

Whether the problem asks for consecutive even numbers or odd numbers, you do not have to do anything different. The pattern is still the same—to get to the next odd or the next even integer, add two.

EXAMPLE \(\PageIndex{13}\)

Find three consecutive even integers whose sum is \(120\).

EXAMPLE \(\PageIndex{14}\)

Find three consecutive even integers whose sum is 102.

\(32, 34, 36\)

EXAMPLE \(\PageIndex{15}\)

Find three consecutive even integers whose sum is \(−24\).

\(−10,−8,−6\)

When a number problem is in a real life context, we still use the same strategies that we used for the previous examples.

EXAMPLE \(\PageIndex{16}\)

A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

According to the National Automobile Dealers Association, the average cost of a car in 2014 was $28,400. This was $1,600 less than six times the cost in 1975. What was the average cost of a car in 1975?

The average cost was $5,000.

EXAMPLE \(\PageIndex{18}\)

US Census data shows that the median price of new home in the U.S. in November 2014 was $280,900. This was $10,700 more than 14 times the price in November 1964. What was the median price of a new home in November 1964?

The median price was $19,300.

Solve Percent Applications

There are several methods to solve percent equations. In algebra, it is easiest if we just translate English sentences into algebraic equations and then solve the equations. Be sure to change the given percent to a decimal before you use it in the equation.

EXAMPLE \(\PageIndex{19}\)

Translate and solve:

• What number is 45% of 84?
• 8.5% of what amount is $4.76?
• 168 is what percent of 112?
• What number is 45% of 80?
• 7.5% of what amount is $1.95?
• 110 is what percent of 88?

ⓐ 36 ⓑ $26 ⓒ \(125 \% \)

EXAMPLE \(\PageIndex{21}\)

• What number is 55% of 60?
• 8.5% of what amount is $3.06?
• 126 is what percent of 72?

ⓐ 33 ⓑ $36 ⓐ \(175 \% \)

Now that we have a problem solving strategy to refer to, and have practiced solving basic percent equations, we are ready to solve percent applications. Be sure to ask yourself if your final answer makes sense—since many of the applications we will solve involve everyday situations, you can rely on your own experience.

EXAMPLE \(\PageIndex{22}\)

The label on Audrey’s yogurt said that one serving provided 12 grams of protein, which is 24% of the recommended daily amount. What is the total recommended daily amount of protein?

EXAMPLE \(\PageIndex{23}\)

One serving of wheat square cereal has 7 grams of fiber, which is 28% of the recommended daily amount. What is the total recommended daily amount of fiber?

EXAMPLE \(\PageIndex{24}\)

One serving of rice cereal has 190 mg of sodium, which is 8% of the recommended daily amount. What is the total recommended daily amount of sodium?

Remember to put the answer in the form requested. In the next example we are looking for the percent.

EXAMPLE \(\PageIndex{25}\)

Veronica is planning to make muffins from a mix. The package says each muffin will be 240 calories and 60 calories will be from fat. What percent of the total calories is from fat?

EXAMPLE \(\PageIndex{26}\)

Mitzi received some gourmet brownies as a gift. The wrapper said each 28% brownie was 480 calories, and had 240 calories of fat. What percent of the total calories in each brownie comes from fat? Round the answer to the nearest whole percent.

EXAMPLE \(\PageIndex{27}\)

The mix Ricardo plans to use to make brownies says that each brownie will be 190 calories, and 76 calories are from fat. What percent of the total calories are from fat? Round the answer to the nearest whole percent.

It is often important in many fields—business, sciences, pop culture—to talk about how much an amount has increased or decreased over a certain period of time. This increase or decrease is generally expressed as a percent and called the percent change .

To find the percent change, first we find the amount of change, by finding the difference of the new amount and the original amount. Then we find what percent the amount of change is of the original amount.

FIND PERCENT CHANGE

\[\text{change}= \text{new amount}−\text{original amount}\]

change is what percent of the original amount?

EXAMPLE \(\PageIndex{28}\)

Recently, the California governor proposed raising community college fees from $36 a unit to $46 a unit. Find the percent change. (Round to the nearest tenth of a percent.)

EXAMPLE \(\PageIndex{29}\)

Find the percent change. (Round to the nearest tenth of a percent.) In 2011, the IRS increased the deductible mileage cost to 55.5 cents from 51 cents.

\(8.8 \% \)

EXAMPLE \(\PageIndex{30}\)

Find the percent change. (Round to the nearest tenth of a percent.) In 1995, the standard bus fare in Chicago was $1.50. In 2008, the standard bus fare was 2.25.

Applications of discount and mark-up are very common in retail settings.

When you buy an item on sale, the original price has been discounted by some dollar amount. The discount rate , usually given as a percent, is used to determine the amount of the discount . To determine the amount of discount, we multiply the discount rate by the original price.

The price a retailer pays for an item is called the original cost . The retailer then adds a mark-up to the original cost to get the list price , the price he sells the item for. The mark-up is usually calculated as a percent of the original cost. To determine the amount of mark-up, multiply the mark-up rate by the original cost.

\[ \begin{align} \text{amount of discount} &= \text{discount rate}· \text{original price} \\ \text{sale price} &= \text{original amount}– \text{discount price} \end{align}\]

The sale price should always be less than the original price.

\[\begin{align} \text{amount of mark-up} &= \text{mark-up rate}·\text{original price} \\ \text{list price} &= \text{original cost}–\text{mark-up} \end{align}\]

The list price should always be more than the original cost.

EXAMPLE \(\PageIndex{31}\)

Liam’s art gallery bought a painting at an original cost of $750. Liam marked the price up 40%. Find

• the amount of mark-up and
• the list price of the painting.

EXAMPLE \(\PageIndex{32}\)

Find ⓐ the amount of mark-up and ⓑ the list price: Jim’s music store bought a guitar at original cost $1,200. Jim marked the price up 50%.

ⓐ $600 ⓑ $1,800

EXAMPLE \(\PageIndex{33}\)

Find ⓐ the amount of mark-up and ⓑ the list price: The Auto Resale Store bought Pablo’s Toyota for $8,500. They marked the price up 35%.

ⓐ $2,975 ⓑ $11,475

Solve Simple Interest Applications

Interest is a part of our daily lives. From the interest earned on our savings to the interest we pay on a car loan or credit card debt, we all have some experience with interest in our lives.

The amount of money you initially deposit into a bank is called the principal , P , and the bank pays you interest, I. When you take out a loan, you pay interest on the amount you borrow, also called the principal.

In either case, the interest is computed as a certain percent of the principal, called the rate of interest , r . The rate of interest is usually expressed as a percent per year, and is calculated by using the decimal equivalent of the percent. The variable t , (for time) represents the number of years the money is saved or borrowed.

Interest is calculated as simple interest or compound interest. Here we will use simple interest.

SIMPLE INTEREST

If an amount of money, P , called the principal, is invested or borrowed for a period of t years at an annual interest rate r , the amount of interest, I , earned or paid is

\[ \begin{array}{ll} I=Prt \; \; \; \; \; \; \; \; \; \; \; \; \text{where} & { \begin{align} I &= \text{interest} \\ P &= \text{principal} \\ r &= \text{rate} \\ t &= \text{time} \end{align}} \end{array}\]

Interest earned or paid according to this formula is called simple interest .

The formula we use to calculate interest is \(I=Prt\). To use the formula we substitute in the values for variables that are given, and then solve for the unknown variable. It may be helpful to organize the information in a chart.

EXAMPLE \(\PageIndex{34}\)

Areli invested a principal of $950 in her bank account that earned simple interest at an interest rate of 3%. How much interest did she earn in five years?

\( \begin{aligned} I & = \; ? \\ P & = \; \$ 950 \\ r & = \; 3 \% \\ t & = \; 5 \text{ years} \end{aligned}\)

\(\begin{array}{ll} \text{Identify what you are asked to find, and choose a} & \text{What is the simple interest?} \\ \text{variable to represent it.} & \text{Let } I= \text{interest.} \\ \text{Write the formula.} & I=Prt \\ \text{Substitute in the given information.} & I=(950)(0.03)(5) \\ \text{Simplify.} & I=142.5 \\ \text{Check.} \\ \text{Is } \$142.50 \text{ a reasonable amount of interest on } \$ \text{ 950?} \; \;\;\;\;\; \;\;\;\;\;\; \\ \text{Yes.} \\ \text{Write a complete sentence.} & \text{The interest is } \$ \text{142.50.} \end{array}\)

EXAMPLE \(\PageIndex{35}\)

Nathaly deposited $12,500 in her bank account where it will earn 4% simple interest. How much interest will Nathaly earn in five years?

He will earn $2,500.

EXAMPLE \(\PageIndex{36}\)

Susana invested a principal of $36,000 in her bank account that earned simple interest at an interest rate of 6.5%.6.5%. How much interest did she earn in three years?

She earned $7,020.

There may be times when we know the amount of interest earned on a given principal over a certain length of time, but we do not know the rate.

EXAMPLE \(\PageIndex{37}\)

Hang borrowed $7,500 from her parents to pay her tuition. In five years, she paid them $1,500 interest in addition to the $7,500 she borrowed. What was the rate of simple interest?

\( \begin{aligned} I & = \; \$ 1500 \\ P & = \; \$ 7500 \\ r & = \; ? \\ t & = \; 5 \text{ years} \end{aligned}\)

Identify what you are asked to find, and choose What is the rate of simple interest? a variable to represent it. Write the formula. Substitute in the given information. Multiply. Divide. Change to percent form. Let r = rate of interest. I = P r t 1,500 = ( 7,500 ) r ( 5 ) 1,500 = 37,500 r 0.04 = r 4 % = r Check. I = P r t 1,500 = ? ( 7,500 ) ( 0.04 ) ( 5 ) 1,500 = 1,500 ✓ Write a complete sentence. The rate of interest was 4%. Identify what you are asked to find, and choose What is the rate of simple interest? a variable to represent it. Write the formula. Substitute in the given information. Multiply. Divide. Change to percent form. Let r = rate of interest. I = P r t 1 ,500 = ( 7,500 ) r ( 5 ) 1,500 = 37,500 r 0.04 = r 4 % = r Check. I = P r t 1 ,500 = ? ( 7,500 ) ( 0.04 ) ( 5 ) 1,500 = 1, 500 ✓ Write a complete sentence. The rate of interest was 4%.

EXAMPLE \(\PageIndex{38}\)

Jim lent his sister $5,000 to help her buy a house. In three years, she paid him the $5,000, plus $900 interest. What was the rate of simple interest?

The rate of simple interest was 6%.

EXAMPLE \(\PageIndex{39}\)

Loren lent his brother $3,000 to help him buy a car. In four years, his brother paid him back the $3,000 plus $660 in interest. What was the rate of simple interest?

The rate of simple interest was 5.5%.

In the next example, we are asked to find the principal—the amount borrowed.

EXAMPLE \(\PageIndex{40}\)

Sean’s new car loan statement said he would pay $4,866,25 in interest from a simple interest rate of 8.5% over five years. How much did he borrow to buy his new car?

\( \begin{aligned} I & = \; 4,866.25 \\ P & = \; ? \\ r & = \; 8.5 \% \\ t & = \; 5 \text{ years} \end{aligned}\)

Identify what you are asked to find, What is the amount borrowed (the principal)? and choose a variable to represent it. Write the formula. Substitute in the given information. Multiply. Divide. Let P = principal borrowed. I = P r t 4,866.25 = P ( 0.085 ) ( 5 ) 4,866.25 = 0.425 P 11,450 = P Check. I = P r t 4,866.25 = ? ( 11,450 ) ( 0.085 ) ( 5 ) 4,866.25 = 4,866.25 ✓ Write a complete sentence. The principal was $11,450. Identify what you are asked to find, What is the amount borrowed (the principal)? and choose a variable to represent it. Write the formula. Substitute in the given information. Multiply. Divide. Let P = principal borrowed. I = P r t 4 ,866.25 = P ( 0.085 ) ( 5 ) 4,866.25 = 0.425 P 11,450 = P Check. I = P r t 4 ,866.25 = ? ( 11,450 ) ( 0.085 ) ( 5 ) 4,866.25 = 4,866.25 ✓ Write a complete sentence. The principal was $11,450.

EXAMPLE \(\PageIndex{41}\)

Eduardo noticed that his new car loan papers stated that with a 7.5% simple interest rate, he would pay $6,596.25 in interest over five years. How much did he borrow to pay for his car?

He paid $17,590.

EXAMPLE \(\PageIndex{42}\)

In five years, Gloria’s bank account earned $2,400 interest at 5% simple interest. How much had she deposited in the account?

She deposited $9,600.

Access this online resource for additional instruction and practice with using a problem solving strategy.

• Begining Arithmetic Problems

Key Concepts

\(\text{change}=\text{new amount}−\text{original amount}\)

\(\text{change is what percent of the original amount?}\)

• \( \begin{align} \text{amount of discount} &= \text{discount rate}· \text{original price} \\ \text{sale price} &= \text{original amount}– \text{discount price} \end{align}\)
• \(\begin{align} \text{amount of mark-up} &= \text{mark-up rate}·\text{original price} \\ \text{list price} &= \text{original cost}–\text{mark-up} \end{align}\)
• If an amount of money, P , called the principal, is invested or borrowed for a period of t years at an annual interest rate r , the amount of interest, I , earned or paid is: \[\begin{aligned} &{} &{} &{I=interest} \nonumber\\ &{I=Prt} &{\text{where} \space} &{P=principal} \nonumber\\ &{} &{\space} &{r=rate} \nonumber\\ &{} &{\space} &{t=time} \nonumber \end{aligned}\]

Practice Makes Perfect

1. List five positive thoughts you can say to yourself that will help you approach word problems with a positive attitude. You may want to copy them on a sheet of paper and put it in the front of your notebook, where you can read them often.

Answers will vary.

2. List five negative thoughts that you have said to yourself in the past that will hinder your progress on word problems. You may want to write each one on a small piece of paper and rip it up to symbolically destroy the negative thoughts.

In the following exercises, solve using the problem solving strategy for word problems. Remember to write a complete sentence to answer each question.

3. There are \(16\) girls in a school club. The number of girls is four more than twice the number of boys. Find the number of boys.

4. There are \(18\) Cub Scouts in Troop 645. The number of scouts is three more than five times the number of adult leaders. Find the number of adult leaders.

5. Huong is organizing paperback and hardback books for her club’s used book sale. The number of paperbacks is \(12\) less than three times the number of hardbacks. Huong had \(162\) paperbacks. How many hardback books were there?

58 hardback books

6. Jeff is lining up children’s and adult bicycles at the bike shop where he works. The number of children’s bicycles is nine less than three times the number of adult bicycles. There are \(42\) adult bicycles. How many children’s bicycles are there?

In the following exercises, solve each number word problem.

7. The difference of a number and \(12\) is three. Find the number.

8. The difference of a number and eight is four. Find the number.

9. The sum of three times a number and eight is \(23\). Find the number.

10. The sum of twice a number and six is \(14\). Find the number.

11 . The difference of twice a number and seven is \(17\). Find the number.

12. The difference of four times a number and seven is \(21\). Find the number.

13. Three times the sum of a number and nine is \(12\). Find the number.

14. Six times the sum of a number and eight is \(30\). Find the number.

15. One number is six more than the other. Their sum is \(42\). Find the numbers.

\(18, \;24\)

16. One number is five more than the other. Their sum is \(33\). Find the numbers.

17. The sum of two numbers is \(20\). One number is four less than the other. Find the numbers.

\(8, \;12\)

18 . The sum of two numbers is \(27\). One number is seven less than the other. Find the numbers.

19. One number is \(14\) less than another. If their sum is increased by seven, the result is \(85\). Find the numbers.

\(32,\; 46\)

20 . One number is \(11\) less than another. If their sum is increased by eight, the result is \(71\). Find the numbers.

21. The sum of two numbers is \(14\). One number is two less than three times the other. Find the numbers.

\(4,\; 10\)

22. The sum of two numbers is zero. One number is nine less than twice the other. Find the numbers.

23. The sum of two consecutive integers is \(77\). Find the integers.

\(38,\; 39\)

24. The sum of two consecutive integers is \(89\). Find the integers.

25. The sum of three consecutive integers is \(78\). Find the integers.

\(25,\; 26,\; 27\)

26. The sum of three consecutive integers is \(60\). Find the integers.

27. Find three consecutive integers whose sum is \(−36\).

\(−11,\;−12,\;−13\)

28. Find three consecutive integers whose sum is \(−3\).

29. Find three consecutive even integers whose sum is \(258\).

\(84,\; 86,\; 88\)

30. Find three consecutive even integers whose sum is \(222\).

31. Find three consecutive odd integers whose sum is \(−213\).

\(−69,\;−71,\;−73\)

32. Find three consecutive odd integers whose sum is \(−267\).

33. Philip pays \($1,620\) in rent every month. This amount is \($120\) more than twice what his brother Paul pays for rent. How much does Paul pay for rent?

34. Marc just bought an SUV for \($54,000\). This is \($7,400\) less than twice what his wife paid for her car last year. How much did his wife pay for her car?

35. Laurie has \($46,000\) invested in stocks and bonds. The amount invested in stocks is \($8,000\) less than three times the amount invested in bonds. How much does Laurie have invested in bonds?

\($13,500\)

36. Erica earned a total of \($50,450\) last year from her two jobs. The amount she earned from her job at the store was \($1,250\) more than three times the amount she earned from her job at the college. How much did she earn from her job at the college?

In the following exercises, translate and solve.

37. a. What number is 45% of 120? b. 81 is 75% of what number? c. What percent of 260 is 78?

a. 54 b. 108 c. 30%

38. a. What number is 65% of 100? b. 93 is 75% of what number? c. What percent of 215 is 86?

39. a. 250% of 65 is what number? b. 8.2% of what amount is $2.87? c. 30 is what percent of 20?

a. 162.5 b. $35 c. 150%

40. a. 150% of 90 is what number? b. 6.4% of what amount is $2.88? c. 50 is what percent of 40?

In the following exercises, solve.

41. Geneva treated her parents to dinner at their favorite restaurant. The bill was $74.25. Geneva wants to leave 16% of the total bill as a tip. How much should the tip be?

42. When Hiro and his co-workers had lunch at a restaurant near their work, the bill was $90.50. They want to leave 18% of the total bill as a tip. How much should the tip be?

43. One serving of oatmeal has 8 grams of fiber, which is 33% of the recommended daily amount. What is the total recommended daily amount of fiber?

44. One serving of trail mix has 67 grams of carbohydrates, which is 22% of the recommended daily amount. What is the total recommended daily amount of carbohydrates?

45. A bacon cheeseburger at a popular fast food restaurant contains 2070 milligrams (mg) of sodium, which is 86% of the recommended daily amount. What is the total recommended daily amount of sodium?

46. A grilled chicken salad at a popular fast food restaurant contains 650 milligrams (mg) of sodium, which is 27% of the recommended daily amount. What is the total recommended daily amount of sodium?

47. The nutrition fact sheet at a fast food restaurant says the fish sandwich has 380 calories, and 171 calories are from fat. What percent of the total calories is from fat?

48. The nutrition fact sheet at a fast food restaurant says a small portion of chicken nuggets has 190 calories, and 114 calories are from fat. What percent of the total calories is from fat?

49. Emma gets paid $3,000 per month. She pays $750 a month for rent. What percent of her monthly pay goes to rent?

50. Dimple gets paid $3,200 per month. She pays $960 a month for rent. What percent of her monthly pay goes to rent?

51. Tamanika received a raise in her hourly pay, from $15.50 to $17.36. Find the percent change.

52. Ayodele received a raise in her hourly pay, from $24.50 to $25.48. Find the percent change.

53. Annual student fees at the University of California rose from about $4,000 in 2000 to about $12,000 in 2010. Find the percent change.

54. The price of a share of one stock rose from $12.50 to $50. Find the percent change.

55. A grocery store reduced the price of a loaf of bread from $2.80 to $2.73. Find the percent change.

−2.5%

56. The price of a share of one stock fell from $8.75 to $8.54. Find the percent change.

57. Hernando’s salary was $49,500 last year. This year his salary was cut to $44,055. Find the percent change.

58. In ten years, the population of Detroit fell from 950,000 to about 712,500. Find the percent change.

In the following exercises, find a. the amount of discount and b. the sale price.

59. Janelle bought a beach chair on sale at 60% off. The original price was $44.95.

a. $26.97 b. $17.98

60. Errol bought a skateboard helmet on sale at 40% off. The original price was $49.95.

In the following exercises, find a. the amount of discount and b. the discount rate (Round to the nearest tenth of a percent if needed.)

61. Larry and Donna bought a sofa at the sale price of $1,344. The original price of the sofa was $1,920.

a. $576 b. 30%

62. Hiroshi bought a lawnmower at the sale price of $240. The original price of the lawnmower is $300.

In the following exercises, find a. the amount of the mark-up and b. the list price.

63. Daria bought a bracelet at original cost $16 to sell in her handicraft store. She marked the price up 45%. What was the list price of the bracelet?

a. $7.20 b. $23.20

64. Regina bought a handmade quilt at original cost $120 to sell in her quilt store. She marked the price up 55%. What was the list price of the quilt?

65. Tom paid $0.60 a pound for tomatoes to sell at his produce store. He added a 33% mark-up. What price did he charge his customers for the tomatoes?

a. $0.20 b. $0.80

66. Flora paid her supplier $0.74 a stem for roses to sell at her flower shop. She added an 85% mark-up. What price did she charge her customers for the roses?

67. Casey deposited $1,450 in a bank account that earned simple interest at an interest rate of 4%. How much interest was earned in two years?

68 . Terrence deposited $5,720 in a bank account that earned simple interest at an interest rate of 6%. How much interest was earned in four years?

69. Robin deposited $31,000 in a bank account that earned simple interest at an interest rate of 5.2%. How much interest was earned in three years?

70. Carleen deposited $16,400 in a bank account that earned simple interest at an interest rate of 3.9% How much interest was earned in eight years?

71. Hilaria borrowed $8,000 from her grandfather to pay for college. Five years later, she paid him back the $8,000, plus $1,200 interest. What was the rate of simple interest?

72. Kenneth lent his niece $1,200 to buy a computer. Two years later, she paid him back the $1,200, plus $96 interest. What was the rate of simple interest?

73. Lebron lent his daughter $20,000 to help her buy a condominium. When she sold the condominium four years later, she paid him the $20,000, plus $3,000 interest. What was the rate of simple interest?

74. Pablo borrowed $50,000 to start a business. Three years later, he repaid the $50,000, plus $9,375 interest. What was the rate of simple interest?

75. In 10 years, a bank account that paid 5.25% simple interest earned $18,375 interest. What was the principal of the account?

76. In 25 years, a bond that paid 4.75% simple interest earned $2,375 interest. What was the principal of the bond?

77. Joshua’s computer loan statement said he would pay $1,244.34 in simple interest for a three-year loan at 12.4%. How much did Joshua borrow to buy the computer?

78. Margaret’s car loan statement said she would pay $7,683.20 in simple interest for a five-year loan at 9.8%. How much did Margaret borrow to buy the car?

Everyday Math

79 . Tipping At the campus coffee cart, a medium coffee costs $1.65. MaryAnne brings $2.00 with her when she buys a cup of coffee and leaves the change as a tip. What percent tip does she leave?

80 . Tipping Four friends went out to lunch and the bill came to $53.75 They decided to add enough tip to make a total of $64, so that they could easily split the bill evenly among themselves. What percent tip did they leave?

Writing Exercises

81. What has been your past experience solving word problems? Where do you see yourself moving forward?

82. Without solving the problem “44 is 80% of what number” think about what the solution might be. Should it be a number that is greater than 44 or less than 44? Explain your reasoning.

83. After returning from vacation, Alex said he should have packed 50% fewer shorts and 200% more shirts. Explain what Alex meant.

84. Because of road construction in one city, commuters were advised to plan that their Monday morning commute would take 150% of their usual commuting time. Explain what this means.

a. After completing the exercises, use this checklist to evaluate your mastery of the objective of this section.

b. After reviewing this checklist, what will you do to become confident for all objectives?

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

Common Core State Standards

How does this relate to 8 th grade and high school math?

• Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
• High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

[FREE] Solving Equations Worksheet (Grade 6 to 8)

Use this worksheet to check your grade 6 to 8 students’ understanding of solving equations. 15 questions with answers to identify areas of strength and support!

How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

• Combine like terms .
• Simplify the equation by using the opposite operation to both sides.
• Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

• Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
• Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
• Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
• Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

• Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
• Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
• Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
• Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
• Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

• Math equations
• Rearranging equations
• How to find the equation of a line
• Solve equations with fractions
• Linear equations
• Writing linear equations
• Substitution
• Identity math
• One step equation

Practice solving equations questions

1. Solve 4x-2=14.

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

• Inequalities
• Types of graph
• Coordinate plane

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Privacy Overview

Chapter 1: Solving Equations and Inequalities

Problem solving, learning objectives.

• Translate words into algebraic expressions and equations
• Define a process for solving word problems
• Apply the steps for solving word problems to distance, rate, and time problems
• Apply the steps for solving word problems to interest rate problems
• Evaluate a formula using substitution
• Rearrange formulas to isolate specific variables
• Identify an unknown given a formula
• Apply the steps for solving word problems to geometry problems
• Use the formula for converting between Fahrenheit and Celsius

Define a Process for Problem Solving

Word problems can be tricky. Often it takes a bit of practice to convert an English sentence into a mathematical sentence, which is one of the first steps to solving word problems. In the table below, words or phrases commonly associated with mathematical operators are categorized. Word problems often contain these or similar words, so it’s good to see what mathematical operators are associated with them.

Some examples follow:

• [latex]x\text{ is }5[/latex]  becomes [latex]x=5[/latex]
• Three more than a number becomes [latex]x+3[/latex]
• Four less than a number becomes [latex]x-4[/latex]
• Double the cost becomes [latex]2\cdot\text{ cost }[/latex]
• Groceries and gas together for the week cost $250 means [latex]\text{ groceries }+\text{ gas }=250[/latex]
• The difference of 9 and a number becomes [latex]9-x[/latex]. Notice how 9 is first in the sentence and the expression

Let’s practice translating a few more English phrases into algebraic expressions.

Translate the table into algebraic expressions:

In this example video, we show how to translate more words into mathematical expressions.

The power of algebra is how it can help you model real situations in order to answer questions about them.

Here are some steps to translate problem situations into algebraic equations you can solve. Not every word problem fits perfectly into these steps, but they will help you get started.

• Read and understand the problem.
• Determine the constants and variables in the problem.
• Translate words into algebraic expressions and equations.
• Write an equation to represent the problem.
• Solve the equation.
• Check and interpret your answer. Sometimes writing a sentence helps.

Twenty-eight less than five times a certain number is 232. What is the number?

Following the steps provided:

• Read and understand: we are looking for a number.
• Constants and variables: 28 and 232 are constants, “a certain number” is our variable because we don’t know its value, and we are asked to find it. We will call it x.
• Translate:  five times a certain number translates to [latex]5x[/latex] Twenty-eight less than five times a certain number translates to [latex]5x-28[/latex] because subtraction is built backward. is 232 translates to [latex]=232[/latex] because “is” is associated with equals.
• Write an equation:  [latex]5x-28=232[/latex]

[latex]\begin{array}{r}5x-28=232\\5x=260\\x=52\,\,\,\end{array}[/latex]

[latex]\begin{array}{r}5\left(52\right)-28=232\\5\left(52\right)=260\\260=260\end{array}[/latex].

In the video that follows, we show another example of how to translate a sentence into a mathematical expression using a problem solving method.

Another type of number problem involves consecutive numbers. Consecutive numbers are numbers that come one after the other, such as 3, 4, 5. If we are looking for several consecutive numbers it is important to first identify what they look like with variables before we set up the equation.

For example, let’s say I want to know the next consecutive integer after 4. In mathematical terms, we would add 1 to 4 to get 5. We can generalize this idea as follows: the consecutive integer of any number, x , is [latex]x+1[/latex]. If we continue this pattern we can define any number of consecutive integers from any starting point. The following table shows how to describe four consecutive integers using algebraic notation.

We apply the idea of consecutive integers to solving a word problem in the following example.

The sum of three consecutive integers is 93. What are the integers?

• Read and understand:  We are looking for three numbers, and we know they are consecutive integers.
• Constants and Variables:  93 is a constant. The first integer we will call x . Second: [latex]x+1[/latex] Third: [latex]x+2[/latex]
• Translate:  The sum of three consecutive integers translates to [latex]x+\left(x+1\right)+\left(x+2\right)[/latex], based on how we defined the first, second, and third integers. Notice how we placed parentheses around the second and third integers. This is just to make each integer more distinct. is 93 translates to [latex]=93[/latex] because is is associated with equals.
• Write an equation:  [latex]x+\left(x+1\right)+\left(x+2\right)=93[/latex]

[latex]x+x+1+x+2=93[/latex]

Combine like terms, simplify, and solve.

[latex]\begin{array}{r}x+x+1+x+2=93\\3x+3 = 93\\\underline{-3\,\,\,\,\,-3}\\3x=90\\\frac{3x}{3}=\frac{90}{3}\\x=30\end{array}[/latex]

• Check and Interpret: Okay, we have found a value for x . We were asked to find the value of three consecutive integers, so we need to do a couple more steps. Remember how we defined our variables: The first integer we will call [latex]x[/latex], [latex]x=30[/latex] Second: [latex]x+1[/latex] so [latex]30+1=31[/latex] Third: [latex]x+2[/latex] so [latex]30+2=32[/latex] The three consecutive integers whose sum is [latex]93[/latex] are [latex]30\text{, }31\text{, and }32[/latex]

There is often a well-known formula or relationship that applies to a word problem. For example, if you were to plan a road trip, you would want to know how long it would take you to reach your destination. [latex]d=rt[/latex] is a well-known relationship that associates distance traveled, the rate at which you travel, and how long the travel takes.

Distance, Rate, and Time

If you know two of the quantities in the relationship [latex]d=rt[/latex], you can easily find the third using methods for solving linear equations. For example, if you know that you will be traveling on a road with a speed limit of [latex]30\frac{\text{ miles }}{\text{ hour }}[/latex] for 2 hours, you can find the distance you would travel by multiplying rate times time or [latex]\left(30\frac{\text{ miles }}{\text{ hour }}\right)\left(2\text{ hours }\right)=60\text{ miles }[/latex].

We can generalize this idea depending on what information we are given and what we are looking for. For example, if we need to find time, we could solve the [latex]d=rt[/latex] equation for t using division:

[latex]d=rt\\\frac{d}{r}=t[/latex]

Likewise, if we want to find rate, we can isolate r using division:

[latex]d=rt\\\frac{d}{t}=r[/latex]

In the following examples you will see how this formula is applied to answer questions about ultra marathon running.

Ultra marathon running (defined as anything longer than 26.2 miles) is becoming very popular among women even though it remains a male-dominated niche sport. Ann Trason has broken twenty world records in her career. One such record was the American River 50-mile Endurance Run which begins in Sacramento, California, and ends in Auburn, California. [1] In 1993 Trason finished the run with a time of 6:09:08.  The men’s record for the same course was set in 1994 by Tom Johnson who finished the course with a time of 5:33:21. [2]

In the next examples we will use the [latex]d=rt[/latex] formula to answer the following questions about the two runners.

What was each runner’s rate for their record-setting runs?

By the time Johnson had finished, how many more miles did Trason have to run?

How much further could Johnson have run if he had run as long as Trason?

What was each runner’s time for running one mile?

To make answering the questions easier, we will round the two runners’ times to 6 hours and 5.5 hours.

Read and Understand:  We are looking for rate and we know distance and time, so we can use the idea: [latex]d=rt\\\frac{d}{t}=r[/latex]

Define and Translate: Because there are two runners, making a table to organize this information helps. Note how we keep units to help us keep track of what how all the terms are related to each other.

Write and Solve:

Trason’s rate:

[latex]\begin{array}{c}d=rt\\\\50\text{ miles }=\text{r}\left(6\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{8.33\text{ miles }}{\text{ hour }}\end{array}[/latex].

(rounded to two decimal places)

Johnson’s rate:

[latex]\begin{array}{c}d=rt\\\\,\,\,\,\,\,\,50\text{ miles }=\text{r}\left(5.5\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{9.1\text{ miles }}{\text{ hour }}\end{array}[/latex]

Check and Interpret:

We can fill in our table with this information.

Now that we know each runner’s rate we can answer the second question.

Here is the table we created for reference:

Read and Understand:  We are looking for how many miles Trason still had on the trail when Johnson had finished after 5.5 hours. This is a distance, and we know rate and time.

Define and Translate:  We can use the formula [latex]d=rt[/latex] again. This time the unknown is d , and the time Trason had run is 5.5 hours.

[latex]\begin{array}{l}d=rt\\\\d=8.33\frac{\text{ miles }}{\text{ hour }}\left(5.5\text{ hours }\right)\\\\d=45.82\text{ miles }\end{array}[/latex].

Have we answered the question? We were asked to find how many more miles she had to run after 5.5 hours.  What we have found is how long she had run after 5.5 hours. We need to subtract [latex]d=45.82\text{ miles }[/latex] from the total distance of the course.

[latex]50\text{ miles }-45.82\text{ miles }=1.48\text{ miles }[/latex]

The third question is similar to the second. Now that we know each runner’s rate, we can answer questions about individual distances or times.

Read and Understand:  The word further implies we are looking for a distance.

Define and Translate:  We can use the formula [latex]d=rt[/latex] again. This time the unknown is d , the time is 6 hours, and Johnson’s rate is [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex]

[latex]\begin{array}{l}d=rt\\\\d=9.1\frac{\text{ miles }}{\text{ hour }}\left(6\text{ hours }\right)\\\\d=54.6\text{ miles }\end{array}[/latex].

Have we answered the question? We were asked to find how many more miles Johnson would have run if he had run at his rate of [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex] for 6 hours.

Johnson would have run 54.6 miles, so that’s 4.6 more miles than than he ran for the race.

Now we will tackle the last question where we are asked to find a time for each runner.

Read and Understand:  we are looking for time, and this time our distance has changed from 50 miles to 1 mile, so we can use

Define and Translate: we can use the formula [latex]d=rt[/latex] again. This time the unknown is t , the distance is 1 mile, and we know each runner’s rate. It may help to create a new table:

We will need to divide to isolate time.

[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=8.33\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{8.33\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.12\text{ hours }=t\end{array}[/latex].

0.12 hours is about 7.2 minutes, so Trason’s time for running one mile was about 7.2 minutes. WOW! She did that for 6 hours!

[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=9.1\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{9.1\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.11\text{ hours }=t\end{array}[/latex].

0.11 hours is about 6.6 minutes, so Johnson’s time for running one mile was about 6.6 minutes. WOW! He did that for 5.5 hours!

Have we answered the question? We were asked to find how long it took each runner to run one mile given the rate at which they ran the whole 50-mile course.  Yes, we answered our question.

Trason’s mile time was [latex]7.2\frac{\text{minutes}}{\text{mile}}[/latex] and Johnsons’ mile time was [latex]6.6\frac{\text{minutes}}{\text{mile}}[/latex]

In the following video, we show another example of answering many rate questions given distance and time.

Simple Interest

In order to entice customers to invest their money, many banks will offer interest-bearing accounts. The accounts work like this: a customer deposits a certain amount of money (called the Principal, or P ), which then grows slowly according to the interest rate ( R , measured in percent) and the length of time ( T , usually measured in months) that the money stays in the account. The amount earned over time is called the interest ( I ), which is then given to the customer.

The simplest way to calculate interest earned on an account is through the formula [latex]\displaystyle I=P\,\cdot \,R\,\cdot \,T[/latex].

If we know any of the three amounts related to this equation, we can find the fourth. For example, if we want to find the time it will take to accrue a specific amount of interest, we can solve for T using division:

[latex]\displaystyle\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,I=P\,\cdot \,R\,\cdot \,T\\\\ \frac{I}{{P}\,\cdot \,R}=\frac{P\cdot\,R\,\cdot \,T}{\,P\,\cdot \,R}\\\\\,\,\,\,\,\,\,\,\,\,\,{T}=\frac{I}{\,R\,\cdot \,T}\end{array}[/latex]

Below is a table showing the result of solving for each individual variable in the formula.

In the next examples, we will show how to substitute given values into the simple interest formula, and decipher which variable to solve for.

If a customer deposits a principal of $2000 at a monthly rate of 0.7%, what is the total amount that she has after 24 months?

Substitute in the values given for the Principal, Rate, and Time.

[latex]\displaystyle\begin{array}{l}I=P\,\cdot \,R\,\cdot \,T\\I=2000\cdot 0.7\%\cdot 24\end{array}[/latex]

Rewrite 0.7% as the decimal 0.007, then multiply.

[latex]\begin{array}{l}I=2000\cdot 0.007\cdot 24\\I=336\end{array}[/latex]

Add the interest and the original principal amount to get the total amount in her account.

[latex] \displaystyle 2000+336=2336[/latex]

She has $2336 after 24 months.

The following video shows another example of finding an account balance after a given amount of time, principal invested, and a rate.

In the following example you will see why it is important to make sure the units of the interest rate match the units of time when using the simple interest formula.

Alex invests $600 at 3.25% monthly interest for 3 years. What amount of interest has Alex earned?

Read and Understand: The question asks for an amount, so we can substitute what we are given into the simple interest formula [latex]I=P\,\cdot \,R\,\cdot \,T[/latex]

Define and Translate:  we know P, R, and T so we can use substitution. R  = 0.0325, P = $600, and T = 3 years. We have to be careful! R is in months, and T is in years.  We need to change T into months because we can’t change the rate—it is set by the bank.

[latex]{T}=3\text{ years }\cdot12\frac{\text{ months }}{ year }=36\text{ months }[/latex]

Substitute the given values into the formula.

[latex]\begin{array}{l} I=P\,\cdot \,R\,\cdot \,T\\\\I=600\,\cdot \,0.035\,\cdot \,36\\\\{I}=756\end{array}[/latex]

We were asked what amount Alex earned, which is the amount provided by the formula. In the previous example we were asked the total amount in the account, which included the principal and interest earned.

Alex has earned $756.

After 10 years, Jodi’s account balance has earned $1080 in interest. The rate on the account is 0.09% monthly. What was the original amount she invested in the account?

Read and Understand: The question asks for the original amount invested, the principal. We are given a length of time in years, and an interest rate in months, and the amount of interest earned.

Define and Translate:  we know I = $1080, R = 0.009, and T = 10 years so we can use [latex]{P}=\frac{I}{{R}\,\cdot \,T}[/latex]

We also need to make sure the units on the interest rate and the length of time match, and they do not. We need to change time into months again.

[latex]{T}=10\text{ years }\cdot12\frac{\text{ months }}{ year }=120\text{ months }[/latex]

Substitute the given values into the formula

[latex]\begin{array}{l}{P}=\frac{I}{{R}\,\cdot \,T}\\\\{P}=\frac{1080}{{0.009}\,\cdot \,120}\\\\{P}=\frac{1080}{1.08}=1000\end{array}[/latex]

We were asked to find the principal given the amount of interest earned on an account.  If we substitute P = $1000 into the formula [latex]I=P\,\cdot \,R\,\cdot \,T[/latex] we get

[latex]I=1000\,\cdot \,0.009\,\cdot \,120\\I=1080[/latex]

Our solution checks out. Jodi invested $1000.

Further Applications of Linear Equations

Formulas come up in many different areas of life. We have seen the formula that relates distance, rate, and time and the formula for simple interest on an investment. In this section we will look further at formulas and see examples of formulas for dimensions of geometric shapes as well as the formula for converting temperature between Fahrenheit and Celsius.

There are many geometric shapes that have been well studied over the years. We know quite a bit about circles, rectangles, and triangles. Mathematicians have proven many formulas that describe the dimensions of geometric shapes including area, perimeter, surface area, and volume.

Perimeter is the distance around an object. For example, consider a rectangle with a length of 8 and a width of 3. There are two lengths and two widths in a rectangle (opposite sides), so we add [latex]8+8+3+3=22[/latex]. Since there are two lengths and two widths in a rectangle, you may find the perimeter of a rectangle using the formula [latex]{P}=2\left({L}\right)+2\left({W}\right)[/latex] where

In the following example, we will use the problem-solving method we developed to find an unknown width using the formula for the perimeter of a rectangle. By substituting the dimensions we know into the formula, we will be able to isolate the unknown width and find our solution.

You want to make another garden box the same size as the one you already have. You write down the dimensions of the box and go to the lumber store to buy some boards. When you get there you realize you didn’t write down the width dimension—only the perimeter and length. You want the exact dimensions so you can have the store cut the lumber for you.

Here is what you have written down:

Perimeter = 16.4 feet Length = 4.7 feet

Can you find the dimensions you need to have your boards cut at the lumber store? If so, how many boards do you need and what lengths should they be?

Read and Understand:  We know perimeter = 16.4 feet and length = 4.7 feet, and we want to find width.

Define and Translate:

Define the known and unknown dimensions:

First we will substitute the dimensions we know into the formula for perimeter:

[latex]\begin{array}{l}\,\,\,\,\,P=2{W}+2{L}\\\\16.4=2\left(w\right)+2\left(4.7\right)\end{array}[/latex]

Then we will isolate w to find the unknown width.

[latex]\begin{array}{l}16.4=2\left(w\right)+2\left(4.7\right)\\16.4=2{w}+9.4\\\underline{-9.4\,\,\,\,\,\,\,\,\,\,\,\,\,-9.4}\\\,\,\,\,\,\,\,7=2\left(w\right)\\\,\,\,\,\,\,\,\frac{7}{2}=\frac{2\left(w\right)}{2}\\\,\,\,\,3.5=w\end{array}[/latex]

Write the width as a decimal to make cutting the boards easier and replace the units on the measurement, or you won’t get the right size of board!

If we replace the width we found, [latex]w=3.5\text{ feet }[/latex] into the formula for perimeter with the dimensions we wrote down, we can check our work:

[latex]\begin{array}{l}\,\,\,\,\,{P}=2\left({L}\right)+2\left({W}\right)\\\\{16.4}=2\left({4.7}\right)+2\left({3.5}\right)\\\\{16.4}=9.4+7\\\\{16.4}=16.4\end{array}[/latex]

Our calculation for width checks out. We need to ask for 2 boards cut to 3.5 feet and 2 boards cut to 4.7 feet so we can make the new garden box.

This video shows a similar garden box problem.

We could have isolated the w in the formula for perimeter before we solved the equation, and if we were going to use the formula many times, it could save a lot of time. The next example shows how to isolate a variable in a formula before substituting known dimensions or values into the formula.

Isolate the term containing the variable, w, from the formula for the perimeter of a rectangle :  

[latex]{P}=2\left({L}\right)+2\left({W}\right)[/latex].

First, isolate the term with  w by subtracting 2 l from both sides of the equation.

[latex] \displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,p\,=\,\,\,\,2l+2w\\\underline{\,\,\,\,\,-2l\,\,\,\,\,-2l\,\,\,\,\,\,\,\,\,\,\,}\\p-2l=\,\,\,\,\,\,\,\,\,\,\,\,\,2w\end{array}[/latex]

Next, clear the coefficient of w by dividing both sides of the equation by 2.

[latex]\displaystyle \begin{array}{l}\underline{p-2l}=\underline{2w}\\\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\,\frac{p-2l}{2}\,\,=\,\,w\\\,\,\,\,\,\,\,\,\,\,\,w=\frac{p-2l}{2}\end{array}[/latex]

You can rewrite the equation so the isolated variable is on the left side.

[latex]w=\frac{p-2l}{2}[/latex]

The area of a triangle is given by [latex] A=\frac{1}{2}bh[/latex] where

A = area b = the length of the base h = the height of the triangle

Remember that when two variables or a number and a variable are sitting next to each other without a mathematical operator between them, you can assume they are being multiplied. This can seem frustrating, but you can think of it like mathematical slang. Over the years, people who use math frequently have just made that shortcut enough that it has been adopted as convention.

In the next example we will use the formula for area of a triangle to find a missing dimension, as well as use substitution to solve for the base of a triangle given the area and height.

Find the base ( b) of a triangle with an area ( A ) of 20 square feet and a height ( h) of 8 feet.

Use the formula for the area of a triangle, [latex] {A}=\frac{{1}}{{2}}{bh}[/latex] .

Substitute the given lengths into the formula and solve for  b.

[latex]\displaystyle \begin{array}{l}\,\,A=\frac{1}{2}bh\\\\20=\frac{1}{2}b\cdot 8\\\\20=\frac{8}{2}b\\\\20=4b\\\\\frac{20}{4}=\frac{4b}{4}\\\\ \,\,\,5=b\end{array}[/latex]

The base of the triangle measures 5 feet.

We can rewrite the formula in terms of b or h as we did with perimeter previously. This probably seems abstract, but it can help you develop your equation-solving skills, as well as help you get more comfortable with working with all kinds of variables, not just x .

Use the multiplication and division properties of equality to isolate the variable b .

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{h}=\frac{bh}{h}\\\\\,\,\,\,\,\,\,\,\frac{2A}{h}=\frac{b\cancel{h}}{\cancel{h}}\end{array}[/latex]

Write the equation with the desired variable on the left-hand side as a matter of convention:

Use the multiplication and division properties of equality to isolate the variable h .

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{b}=\frac{bh}{b}\\\\\,\,\,\,\,\,\,\,\frac{2A}{b}=\frac{h\cancel{b}}{\cancel{b}}\end{array}[/latex]

Temperature

Let’s look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.

[latex]C=\left(F-32\right)\cdot \frac{5}{9}[/latex]

Given a temperature of [latex]12^{\circ}{C}[/latex], find the equivalent in [latex]{}^{\circ}{F}[/latex].

Substitute the given temperature in[latex]{}^{\circ}{C}[/latex] into the conversion formula:

[latex]12=\left(F-32\right)\cdot \frac{5}{9}[/latex]

Isolate the variable F to obtain the equivalent temperature.

[latex]\begin{array}{r}12=\left(F-32\right)\cdot \frac{5}{9}\\\\\left(\frac{9}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\\left(\frac{108}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\21.6=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+32}\,\,\,\,\,\,\,\,\,\,\,\,\\\\53.6={}^{\circ}{F}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

As with the other formulas we have worked with, we could have isolated the variable F first, then substituted in the given temperature in Celsius.

Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F.

To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by [latex] \displaystyle \frac{9}{5}[/latex].

[latex]\begin{array}{l}\\\,\,\,\,\left(\frac{9}{5}\right)C=\left(F-32\right)\left(\frac{5}{9}\right)\left(\frac{9}{5}\right)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\frac{9}{5}C=F-32\end{array}[/latex]

Add 32 to both sides.

[latex]\begin{array}{l}\frac{9}{5}\,C+32=F-32+32\\\\\frac{9}{5}\,C+32=F\end{array}[/latex]

[latex]F=\frac{9}{5}C+32[/latex]

Think About It

Express the formula for the surface area of a cylinder, [latex]s=2\pi rh+2\pi r^{2}[/latex], in terms of the height, h .

In this example, the variable h is buried pretty deeply in the formula for surface area of a cylinder. Using the order of operations, it can be isolated. Before you look at the solution, use the box below to write down what you think is the best first step to take to isolate h .

[latex]\begin{array}{r}S\,\,=2\pi rh+2\pi r^{2} \\ \underline{-2\pi r^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,-2\pi r^{2}}\\S-2\pi r^{2}\,\,\,\,=\,\,\,\,2\pi rh\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Next, isolate the variable h by dividing both sides of the equation by [latex]2\pi r[/latex].

[latex]\begin{array}{r}\frac{S-2\pi r^{2}}{2\pi r}=\frac{2\pi rh}{2\pi r} \\\\ \frac{S-2\pi r^{2}}{2\pi r}=h\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

• "Ann Trason." Wikipedia. Accessed May 05, 2016. https://en.wikipedia.org/wiki/Ann_Trason . ↵
•  "American River 50 Mile Endurance Run." Wikipedia. Accessed May 05, 2016. https://en.wikipedia.org/wiki/American_River_50_Mile_Endurance_Run . ↵
• Writing Algebraic Expressions. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/uD_V5t-6Kzs . License : CC BY: Attribution
• Write and Solve a Linear Equations to Solve a Number Problem (1). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/izIIqOztUyI . License : CC BY: Attribution
• Write and Solve a Linear Equations to Solve a Number Problem (Consecutive Integers). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/S5HZy3jKodg . License : CC BY: Attribution
• Screenshot: Ann Trason Trail Running. Authored by : Lumen Learning. License : CC BY: Attribution
• Problem Solving Using Distance, Rate, Time (Running). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/3WLp5mY1FhU . License : CC BY: Attribution
• Simple Interest - Determine Account Balance (Monthly Interest). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/XkGgEEMR_00 . License : CC BY: Attribution
• Simple Interest - Determine Interest Balance (Monthly Interest). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/mRV5ljj32Rg . License : CC BY: Attribution
• Simple Interest - Determine Principal Balance (Monthly Interest). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/vbMqN6lVoOM . License : CC BY: Attribution
• Find the Width of a Rectangle Given the Perimeter / Literal Equation. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/jlxPgKQfhQs . License : CC BY: Attribution
• Find the Base of a Triangle Given Area / Literal Equation. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/VQZQvJ3rXYg . License : CC BY: Attribution
• Convert Celsius to Fahrenheit / Literal Equation. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/DRydX8V-JwY . License : CC BY: Attribution
• Ann Trason. Provided by : Wikipedia. Located at : https://en.wikipedia.org/wiki/Ann_Trason . License : CC BY-SA: Attribution-ShareAlike
• American River 50 Mile Endurance Run. Provided by : Wikipedia. Located at : https://en.wikipedia.org/wiki/American_River_50_Mile_Endurance_Run . License : CC BY-SA: Attribution-ShareAlike

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Problem Solving: Use a Formula

What Is It?

Using a Formula is a problem-solving strategy that students can use to find answers to math problems involving geometry , percents , measurement , or algebra. To solve these problems, students must choose the appropriate formula and substitute data in the correct places of a formula. The following problem would be best solved using a formula:

If it is 46 degrees Celsius, how many degrees Fahrenheit is it?

Students can use the formula F = 1.8C + 32 to find the solution.

Why Is It Important?

Using a Formula is a problem-solving strategy that can be used for problems that involve converting units or measuring geometric objects. Also, real-world problems such as tipping in a restaurant, finding the price of a sale item, and buying enough paint for a room all involve using formulas.

How Can You Make It Happen?

Introduce a problem to students that requires them to use a formula to solve the problem. For example:

A rectangle has an area of 40 square meters. If the perimeter of the rectangle is 26 meters, what are the length and the width of the rectangle?

Understand the Problem

Demonstrate that the first step to solving the problem is understanding it. This involves identifying the key pieces of information needed to find the answer. This may require students to read the problem several times or put the problem into their own words. Here are a few formulas that students can use to solve this problem:

A = L x W 40 = L x W

2W + 2L = P 2W + 2L = 26

(Note: L and W can be interchanged in this problem.)

Choose a Strategy

The strategy of Using a Formula can be used in situations where measurements are required to find the solution.

Solve the Problem

I know 40 is a product of 2 and 20, 4 and 10, or 5 and 8.

I looked at the possible answers that would fit the formula for the area of the rectangle. Then I put the numbers into the formula for the perimeter. The numbers 5 and 8 are the two numbers that work for both formulas.

Read the problem again to be sure the question was answered.

I found the length and width of the rectangle, 5 meters and 8 meters.

Check the math to be sure it is correct.

A = L x W 40 = L x W 40 = 8 x 5 P = 2L + 2W 26 = 2L + 2W 26 + 2(8) + 2(5)

Determine if the best strategy was chosen for this problem, or if there was another way to solve the problem.

Using a formula was a good strategy to use for this problem.

The last step is explaining how you found the answer. Demonstrate how to write a paragraph describing the steps taken and how decisions were made throughout the process.

Students should explain their answer and the process they went through to solve it. It is important for students to talk or write about their thinking.

I knew the formula for area and perimeter, so I wrote down the formulas. A = L x W P = 2L + 2W I added the information I knew, which was the area and perimeter of this rectangle. 40 = L x W 26 = 2L + 2W Then I wrote down the numbers that could be the length and width if the area is 40 square meters. I know 40 is a product of 2 and 20, 4 and 10, or 5 and 8. I took those possible numbers and used them with the formula for perimeter. The numbers that did not also fit into this formula, I eliminated. I eliminated 2 and 20 as well as 4 and 10, since those numbers did not work in the perimeter formula. I was left with the numbers 8 and 5. Since they are interchangeable in this problem, I assigned the width as 5 meters and the length as 8 meters.

Guided Practice

Have students try to solve the following problem using the strategy of Using a Formula.

In an isosceles triangle, the unequal side measures half the length of one of the two equal sides. What is the perimeter if the length of the unequal side is 5 cm? What is the length of the sides if the perimeter is 80 cm?

Have students work in pairs, in groups, or individually to solve this problem. They should be able to tell or write about how they found the answer and justify their reasoning.

How Can You Stretch Students' Thinking?

Math problems requiring formulas can be simple, with few criteria needed to solve them, or they can be multidimensional, requiring charts or tables to organize students' thinking. Including more than one formula in a problem, or having multiple correct answers to a problem will help stretch this strategy.

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• Formulas for Problem Solving

Key Questions

A few examples...

Explanation:

I will assume that you mean things like common identities and the quadratic formula . Here are just a few:

Difference of squares identity

#a^2-b^2 = (a-b)(a+b)#

Deceptively simple, but massively useful.

For example:

#a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2# #color(white)(a^4+b^4) = (a^2+b^2)^2 - (sqrt(2)ab)^2# #color(white)(a^4+b^4) = ((a^2+b^2) - sqrt(2)ab)((a^2+b^2) +sqrt(2)ab)# #color(white)(a^4+b^4) = (a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2)#

Difference of cubes identity

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

Sum of cubes identity

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

Quadratic formula

Very useful to know, better if you know how to derive it:

The zeros of #ax^2+bx+c# are given by:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

Pythagoras theorem

If a right angled triangle has legs of length #a, b# and hypotenuse of length #c# then:

#c^2 = a^2+b^2#

This is also very useful in trigonometric form. If we have an angle #theta# in a right-angled triangle, then we call the side nearest #theta# , the #"adjacent"# side, the side opposite it the #"opposite"# side and the hypotenuse the #"hypotenuse"# .

#"hypotenuse"^2 = "adjacent"^2 + "opposite"^2#

Dividing both sides by #"hypotenuse"^2# , we get:

#1 = ("adjacent"/"hypotenuse")^2 + ("opposite"/"hypotenuse")^2#

#1 = cos^2 theta + sin^2 theta#

Then dividing both sides by #cos^2 theta# we find:

#sec^2 theta = 1 + tan^2 theta#

Binomial theorem

#(a+b)^n = sum_(k=0)^n ((n), (k)) a^(n-k) b^k#

where #((n), (k)) = (n!)/((n-k)! k!)#

#(x+1)^4 = x^4+4x^3+6x^2+4x+1#

Using Pythagoras' Theorem, we know #A^2 + B^2 = C^2#

#A^2 = 9# , #B^2 = 16 implies A^2 + B^2 = 9 + 16 = 25#

SInce #C^2 = A^2 + B^2 = 25# , you know that

#C = sqrt(C^2) = sqrt(25) = 5#

You simply find out what value goes with what part of the formula, and do this for each, and then work out the formula as normal.

Python Numerical Methods

This notebook contains an excerpt from the Python Programming and Numerical Methods - A Guide for Engineers and Scientists , the content is also available at Berkeley Python Numerical Methods .

The copyright of the book belongs to Elsevier. We also have this interactive book online for a better learning experience. The code is released under the MIT license . If you find this content useful, please consider supporting the work on Elsevier or Amazon !

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How To Solve Differential Equations Using ODE45 In Matlab

ODE45 is MATLAB's go-to solver for non-stiff differential equations. In this article, developers will find a concise exploration of its application, from setting initial conditions to analyzing output results. Whether you're new to MATLAB, this guide offers clarity on ODE45's utility.

ODE45 is a go-to function in Matlab for solving ordinary differential equations. It's efficient, reliable, and widely used in various engineering and scientific applications. In this article, you'll get hands-on experience and practical insights to make the most out of this essential tool. Whether you're a seasoned developer or just getting started, you'll find valuable tips to enhance your coding skills.

💡KEY INSIGHTS

• The `ode45` function in MATLAB is essential for solving ordinary differential equations using numerical methods.
• Understanding the usage of function handles and time spans in `ode45` is critical for accurate and efficient problem-solving.
• Customizing options and parameters in `ode45` can significantly influence the accuracy and speed of the solutions.
• Integrating `ode45` with MATLAB's plotting functions enables visualization of the solutions, enhancing analysis and interpretation.

Understanding ODE45 Function Syntax

Setting up initial conditions, choosing time spans, implementing ode functions, running the solver, analyzing output results, plotting solutions, error handling and debugging, optimizing performance, frequently asked questions, function handle, time span and initial conditions.

The ODE45 function in Matlab is a versatile tool for solving ordinary differential equations (ODEs). The basic syntax is straightforward and consists of three main components: the function handle, time span, and initial conditions.

The function handle is essentially a pointer to the function that defines the ODE. This function should accept two arguments: time t and state y , and return the derivatives.

The time span and initial conditions are specified as vectors. The time span is a two-element vector [t_start, t_end] , and the initial conditions are a column vector.

Once you've set up the function handle, time span, and initial conditions, you can run the solver.

Scalar Vs Vector

Complex initial conditions, importance of accurate initial conditions.

The initial conditions are a crucial part of solving any ODE. In Matlab's ode45 , these conditions are specified as a column vector. The vector's length should match the number of equations in your ODE system.

For a single equation , a scalar value suffices. However, for a system of equations , you'll need a column vector where each element represents the initial condition for a corresponding equation.

Sometimes, you may encounter complex initial conditions involving imaginary numbers. Matlab handles these seamlessly.

Accurate initial conditions are essential for obtaining a reliable solution . Even a small error can lead to significant deviations in the final output. Therefore, always double-check your initial conditions before running the solver.

By understanding how to properly set up initial conditions, you'll ensure that your ODE solutions are both accurate and meaningful.

Short Vs Long Time Spans

Non-uniform time points, time span sensitivity.

The time span is another critical parameter when using ode45 . It defines the interval over which the differential equation will be solved. The time span is specified as a two-element vector [t_start, t_end] .

The length of the time span can affect the accuracy and computational time. Shorter time spans are quicker to compute but may not capture the system's behavior adequately. Longer time spans offer a more comprehensive view but may require more computational resources.

In some cases, you might want to specify non-uniform time points where the solution should be evaluated. You can do this by providing a vector with more than two elements.

Be cautious when choosing the time span, as it can affect the stability of the solution. Some systems are sensitive to the time span, and an inappropriate choice can lead to erroneous results.

By carefully selecting the time span, you can balance computational efficiency with solution accuracy, ensuring that your results are both reliable and insightful.

Function Arguments

Systems of equations, parameterized functions.

The core of any ODE solver is the ODE function itself. This function defines the differential equation you want to solve. In Matlab, you'll typically create a separate function file for this purpose.

The function arguments t and y are standard for any ODE function in Matlab. The variable t represents time, and y represents the state or value of the function at that time.

When dealing with a system of equations , the ODE function should return a column vector of derivatives, one for each equation in the system.

Sometimes you may need to include additional parameters in your ODE function. You can achieve this using anonymous functions or function handles.

By understanding how to implement ODE functions effectively, you can tackle a wide range of problems with confidence and precision.

Output Variables

Solver options, handling events.

Once you've defined your ODE function, initial conditions, and time span, the next step is to run the solver . The ode45 function in Matlab makes this process straightforward.

The output variables t and y store the time points and corresponding solutions, respectively. These can be used for further analysis or plotting.

Matlab allows you to customize the solver's behavior using solver options , set via the odeset function. These options can control tolerances, step sizes, and more.

You can also specify event functions to stop the solver when certain conditions are met, like reaching a specific state value.

By understanding how to run the solver effectively, you can ensure that your ODE solutions are both accurate and computationally efficient.

Data Inspection

Error analysis, data visualization, exporting results.

Once the solver has run, the next step is analyzing the output . Matlab provides several tools to help you understand the results, starting with basic plotting functions.

Before diving into more complex analyses, it's good practice to inspect the data . Simple commands can provide quick insights into your results.

Understanding the accuracy of your solution is crucial. You can compare the numerical solution to an analytical one, if available, to assess error.

Advanced data visualization techniques like 3D plots or contour maps can offer deeper insights into system behavior.

Finally, you may need to export the results for further analysis or for sharing with others. Matlab supports various formats like CSV, MAT, or even direct export to Excel.

By effectively analyzing the output results, you can derive meaningful conclusions and make informed decisions for subsequent steps in your project.

Multiple Solutions

Customizing plots, interactive plots.

Visualizing your ODE solutions is often the most intuitive way to understand the results . Matlab's plot function is the simplest tool for this.

When you have multiple solutions to compare, you can plot them on the same graph for easier analysis.

Matlab offers a variety of options to customize your plots , from line styles to marker types.

For more complex data sets, subplots can be useful to display multiple graphs in a single figure.

For a more dynamic view, you can use interactive plots like zoom and pan to explore the details of your solution.

By effectively plotting your solutions, you can gain valuable insights into the behavior of the system you're studying, making it easier to interpret and present your findings.

Common Errors

Debugging techniques, logging errors.

When working with ODE solvers, you're likely to encounter errors or issues . Matlab provides built-in functions like try and catch to handle these gracefully.

Some common errors include incorrect function handles, undefined variables, or incompatible dimensions.

Matlab's debugger is a powerful tool for identifying and fixing issues in your code.

Sometimes, Matlab will issue warnings instead of errors. These are non-fatal but should be addressed.

For long-running simulations, logging errors can be useful for post-analysis.

By employing effective error handling and debugging techniques, you can ensure a smoother development process and more reliable results.

Vectorization

Preallocation, parallel computing, profiling code.

When solving ODEs, performance optimization can save you both time and computational resources. One way to improve performance is by adjusting the solver options.

Vectorized operations can significantly speed up your code by eliminating the need for loops.

Memory preallocation is another technique that can improve speed by reducing the time spent on dynamic memory allocation.

For computationally intensive tasks, parallel computing can be a game-changer.

To identify bottlenecks, Matlab provides a code profiler that measures the time taken by each function and line of code.

By applying these optimization techniques, you can make your ODE solving process more efficient and effective.

Can I Use ODE45 for Stiff Equations?

ODE45 is designed for non-stiff equations. For stiff equations, it's better to use solvers like ode15s or ode23s .

What Are the Default Tolerances for ODE45?

The default relative tolerance is 1e-3 , and the default absolute tolerance is 1e-6 . You can change these using the odeset function.

What's the Difference Between ode45 and ode23 ?

ode45 uses a 4th and 5th order Runge-Kutta method, making it more accurate but potentially slower. ode23 uses a 2nd and 3rd order method, which may be faster but less accurate.

How Do I Stop the Solver at a Specific Condition?

You can use an event function to stop the solver when a specific condition is met. Define the event function and pass it to odeset under the 'Events' field.

How Can I Solve a System of ODEs?

To solve a system of ODEs, define your function to return a vector of derivatives. The initial condition should also be a vector corresponding to each equation.

Let’s test your knowledge!

What is the default relative tolerance in ODE45?

Continue learning with these matlab guides.

• How To Use MATLAB Indexing Efficiently
• How To Create And Manipulate A Matlab Table
• How To Manipulate MATLAB Colors For Enhanced Visualization
• How To Use Fprintf Matlab For Effective Text And Data Formatting
• How To Use Matlab Find Function Effectively

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Forming & Solving Equations - Year 11 PDF Download

Cie igcse maths: extended, revision notes.

Syllabus Edition

First teaching 2023

First exams 2025

Forming Equations

Before solving an equation you may need to form it from the information given in the question.

Paraphrasing Information

Take the given information and restate it in your own words to create the equation.

Elaboration with Examples

Provide detailed explanations and examples to ensure understanding.

Bullet Points Importance

Use bullet points to highlight key details and make the content more digestible.

Clarity in Expression Formation

Ensure that the equation formed is clear and accurately represents the given information.

Understanding Expressions in Algebra

• An expression in algebra is a mathematical statement that doesn't have an equals sign, like "2x + 3" or "y - 5".
• Expressions are useful for representing unknown values in mathematical problems.
• When an unknown value needs to be represented, it can be denoted by a letter, such as "x" or "y".
• Common phrases can be translated into expressions by assigning variables to the unknowns.
• Using brackets can help indicate the correct order of operations within an expression.
• For example, "something add 1 then multiplied by 3" can be written as "3(something + 1)".
• It's important to choose the smallest value possible to represent a letter in an expression for simplicity.
• If Adam is 10 years younger than Barry, and we represent Adam's age as "x", then Barry's age can be represented as "x + 10".
• If Adam's age is half of Barry's age, and we represent Adam's age as "y", then Barry's age can be represented as "2y".

Illustrative Examples

Understanding Age Relationships: Adam and Barry

• If Adam is 10 years younger than Barry, then Barry is 10 years older than Adam. Represent Adam's age as x , then Barry's age is x + 10 . This simplifies algebraic expressions.
• If Adam's age is half of Barry's age, then Barry's age is double Adam's age. So if Adam's age is x , then Barry's age is 2x . This simplifies the algebraic expressions further.

How to Form an Equation?

To create an equation based on the given information, follow these steps:

• Identify the relationship between Adam and Barry's ages.
• Assign a variable to represent one of their ages.
• Express the other age in terms of the variable.
• Set up the equation based on the given conditions.

An Equation Explained

• An equation is a mathematical expression containing an equals sign that can be solved to find the value of the variable.
• To solve an equation, you first create an expression and set it equal to a specific value or another expression.
• Addition: sum, total, more than, increase, etc.
• Subtraction: difference, less than, decrease, etc.
• Multiplication: product, lots of, times as many, double, triple, etc.
• Division: shared, split, grouped, halved, quartered, etc.
• For instance, in a scenario where Adam is 10 years younger than Barry and their combined age is 25, you can determine their individual ages by setting up and solving an equation.

Understanding Equations

• An equation is a fundamental mathematical concept involving an equals sign for solving unknown values.
• When encountering equations, it's essential to create expressions and equate them to solve for variables.

Problem Solving with Equations

• If we have two individuals, Adam and Barry, with a specific age relationship, we can represent their ages using equations.
• Let's say Adam is 10 years younger than Barry and the sum of their ages is 25. To find their ages, we can set up the following equation: Adam's age + Barry's age = 25 .

• We can then solve this equation to determine Adam's and Barry's ages.

Simultaneous Equations

• Sometimes, we encounter situations where we have two unknown values (x and y) and need to create two simultaneous equations to solve for them.
• For instance, consider a scenario where the prices of a child's ticket and an adult's ticket at a theatre are given.

• We can write equations based on the given information to represent the relationship between the prices of adult and child tickets.
• a) Adult price = 2 x Child price

• b) Another representation: Adult price = Child price + $7

• c) Total cost equation: 3 x Child price + 2 x Adult price = $45

Forming Equations from Shapes

Many questions involve having to form and solve equations from information given about things relating to shapes, like lengths or angles.

• Rectangles:
• To find the perimeter of a rectangle, add up all the sides: Perimeter = 2 * (length + width)
• To find the area of a rectangle, multiply the length by the width: Area = length * width
• The perimeter of a square is four times the length of one side: Perimeter = 4 * side length
• The area of a square is the side length squared: Area = side length * side length
• To find the perimeter of a triangle, add the lengths of all three sides.
• The area of a triangle is calculated using the formula: Area = 0.5 * base * height
• The circumference of a circle is given by: Circumference = 2 * π * radius
• The area of a circle is: Area = π * radius^2

Strategies for Solving Geometry Problems

• When faced with a geometry problem, start by carefully reading the question to determine if it involves area, perimeter, or angles.
• If a diagram is not provided, it is advisable to quickly sketch one as it can provide valuable visual cues.
• Include any information provided in the question onto the diagram, usually in the form of expressions involving one or two variables.
• For perimeter-related questions, identify which sides have equal lengths and sum them up. Consider the shape's properties to determine side equality.
• In shapes like squares or rhombuses, all sides are equal; in rectangles or parallelograms, opposite sides are equal. For triangles, check for sides of equal length.
• For questions involving area, jot down the necessary formula for the area of the given shape. If the shape is uncommon, break it down into simpler shapes to calculate their areas.
• Remember, a regular polygon has all sides of equal length. For example, a regular pentagon with side length 2x - 1 has 5 equal sides, making its perimeter 5(2x - 1).
• If dealing with a circle or a part of it, use the constant π consistently instead of multiplying by it, preventing long decimals in your calculations.

Key Steps to Tackle Geometry Problems

• Always begin by analyzing the question to discern the type of geometric problem you are solving: area, perimeter, or angles.
• Sketch a diagram promptly if one is not provided to enhance your understanding of the problem.
• Transfer any relevant data from the question onto your diagram, typically involving algebraic expressions with variables.
• When addressing perimeter queries, identify and combine sides of equal length based on the characteristics of the shape. Remember the unique properties of different geometric figures.

Geometric Shapes and Formulas

• In a square or rhombus, all four sides are equal
• In a rectangle or parallelogram, opposite sides are equal
• If a triangle is given, check if any of the sides are equal in length

Calculating Area of Shapes

• If the question involves area, write down the necessary formula for the area of that shape
• If it is an uncommon shape, consider splitting it into two or more common shapes to find the areas
• Remember to split the length and width accordingly for calculations
• Remember to split the length and width accordingly

Special Considerations for Regular Polygons

• Regular polygons have all sides of equal length
• For example, a regular pentagon with side length 2x – 1 has 5 equal sides, making its perimeter 5(2x – 1)
• If one of the shapes is a circle or part of a circle, use π throughout rather than dealing with long decimals

Forming Equations with Angles in 2D Shapes

#Paraphrase the information given, elaborate, and explain with examples wherever required. Present all points in bullet format with proper headings. Ensure language is easy to understand.

Key Concepts for Understanding Geometric Shapes and Angles

• If no diagram is provided, it is beneficial to draw a quick sketch to aid in visualization.
• Include any relevant information from the question in the diagram to enhance understanding.
• The provided information often involves expressions with one or two variables.
• Examine the angles within the shape to identify sides with equal lengths.
• For triangles, consider the equality of angles - an isosceles triangle has two equal angles, while an equilateral triangle has three equal angles.

Angles in Various Geometric Figures

• In parallel lines, consider alternative, corresponding, and co-interior angles.
• In a parallelogram or rhombus, opposite angles are equal, and all four angles sum up to 360 degrees.
• A kite exhibits one pair of equal opposite angles.
• An isosceles triangle features two equal angles, while an equilateral triangle has three equal angles.
• For any polygon with n sides, the sum of its angles is given by 180 degrees multiplied by (n - 2).
• Regular polygons have all angles equal, while irregular polygons are assumed to have different angles unless specified otherwise.
• Be attentive to key details that can provide insights into angle properties, such as symmetry lines in a trapezium indicating pairs of equal angles.

Sum of Angles in Polygons

• For a polygon of n sides, the sum of the angles will be 180°×(n - 2). Remember that a regular polygon means all the angles are equal.
• If a question involves an irregular polygon, assume all the angles are different unless told otherwise.
• Look out for key information that can give more information about the angles.
• For example, a trapezium "with a line of symmetry" will have two pairs of equal angles.

Angles in Polygons Details

• For a polygon of n sides, the sum of the angles will be 180°×(n - 2).
• Remember that a regular polygon means all the angles are equal.

Forming Equations for 3D Shapes

• How do I form an equation involving the surface area or volume of a 3D shape?
• Paraphrase the information given, elaborate, and explain with examples wherever required.
• Give all points in bullet format with proper headings.
• Make sure not more than 2 headings have an HTML tag H7 and not more than 4 subheadings have an HTML tag H8; otherwise, make the headings bold.
• Make sure everything is paraphrased and in detail and keep the language easy to understand.

Optimizing Learning in Geometry: Surface Area and Volume

• Read the question carefully to determine if it involves surface area or volume.
• Mixing these up is a common error observed in GCSE exams.
• If a diagram is not provided, consider sketching a quick one for clarity.
• Include any provided information from the question in your diagram.
• This information typically includes expressions involving one or two variables.
• Examine the properties of the given shape to identify sides with equal lengths.
• In a cube, all sides are equal; all prisms share the same cross-sectional shape at the front and back.
• Pyramids commonly incorporate the factor 1/3 in their formulae.
• If the problem pertains to volume, jot down the requisite formula for the shape's area.
• If the shape is unconventional, the exam question will provide the necessary formula. Sub in expressions for side lengths into the formula, ensuring bracketed expressions are correctly substituted.
• If surface area is the focus:
• Write down the number of faces the shape has and identify any identical ones.
• Determine the 2D shape of each face, noting down the area formula for each.
• Substitute the given expressions into the appropriate formula for each face, ensuring accurate dimension identification. Adjust expressions as needed.
• Compute the total area by summing up all expressions, ensuring each face is considered, even hidden ones in the diagram.

Step-by-Step Guide for Calculating the Perimeter of a Rectangle

• Write down the number of faces the shape has and identify if any are the same.
• Identify the 2D shape of each face and note down the formula for the area of each one.
• Substitute the given expressions into the formula for each face, ensuring the correct dimensions are used.
• You may need to add or subtract some expressions.
• Add the expressions together, ensuring one expression for each face is accounted for.
• Remember to consider any hidden faces in the diagram.
• Use a pencil to carefully annotate the diagrams.
• Most of the work for a question may involve the diagram itself.
• Always read the question carefully - for instance, find the area if asked for, not the perimeter; find the volume if it asks for surface area, etc.

Calculation Example for a Rectangle

Consider a rectangle with a length of 3 cm and a width of 1 cm.

The rectangle's perimeter is given as 22 cm.

The formula for the perimeter of a rectangle is P = 2(length) + 2(width).

Substitute the values: P = 2(3) + 2(1).

Expanding and Simplifying Equations

• Expand the brackets: When we have an expression like P = 2(3x - 1) + 2(2x - 5) , we distribute the values inside the brackets to simplify it. The expanded form of this expression is 6x - 2 + 4x - 10 .
• Simplify the expression: By combining like terms, the simplified form of the expression becomes 10x - 12 .
• Setting it equal to the perimeter: If the perimeter is given as 22, we equate the expression to 22 and solve for x.
• Solving for x: By solving the equation 10x - 12 = 22 , we find that x = 3 .

Further Steps:

• Adding 4 to both sides: To isolate x, we add 4 to both sides of the equation 10x - 12 = 22 . This results in the equation 10x = 26 .
• Dividing both sides by 5: By dividing both sides by 5, we find the value of x to be x = 3 .

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Unit 5: Quadratic functions and equations

About this unit, solving quadratics by taking the square root.

• Solving quadratics by taking square roots (Opens a modal)
• Solving quadratics by taking square roots examples (Opens a modal)
• Quadratics by taking square roots Get 3 of 4 questions to level up!

Vertex form

• Vertex form introduction (Opens a modal)
• Graphing quadratics: vertex form (Opens a modal)
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Solving quadratics by factoring

• Solving quadratics by factoring (Opens a modal)
• Solving quadratics by factoring: leading coefficient ≠ 1 (Opens a modal)
• Solving quadratics using structure (Opens a modal)
• Quadratic equations word problem: triangle dimensions (Opens a modal)
• Quadratic equations word problem: box dimensions (Opens a modal)
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The quadratic formula

• The quadratic formula (Opens a modal)
• Understanding the quadratic formula (Opens a modal)
• Worked example: quadratic formula (example 2) (Opens a modal)
• Worked example: quadratic formula (negative coefficients) (Opens a modal)
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Completing the square

• Completing the square (Opens a modal)
• Worked example: Completing the square (intro) (Opens a modal)
• Worked example: Rewriting expressions by completing the square (Opens a modal)
• Worked example: Rewriting & solving equations by completing the square (Opens a modal)
• Solve by completing the square: Integer solutions (Opens a modal)
• Solve by completing the square: Non-integer solutions (Opens a modal)
• Worked example: completing the square (leading coefficient ≠ 1) (Opens a modal)
• Solving quadratics by completing the square: no solution (Opens a modal)
• Proof of the quadratic formula (Opens a modal)
• Solving quadratics by completing the square (Opens a modal)
• Completing the square (intro) Get 3 of 4 questions to level up!
• Solve equations by completing the square Get 3 of 4 questions to level up!
• Completing the square Get 3 of 4 questions to level up!

Forms and features of quadratic functions

• Forms & features of quadratic functions (Opens a modal)
• Vertex & axis of symmetry of a parabola (Opens a modal)
• Finding features of quadratic functions (Opens a modal)
• Creativity break: What can we do to expand our creative skills? (Opens a modal)
• Interpret quadratic models: Factored form (Opens a modal)
• Interpret quadratic models: Vertex form (Opens a modal)
• Graphing quadratics review (Opens a modal)
• Features of quadratic functions Get 3 of 4 questions to level up!
• Graph parabolas in all forms Get 3 of 4 questions to level up!
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Mathematics > Analysis of PDEs

Title: energy solutions of the cauchy-dirichlet problem for fractional nonlinear diffusion equations.

Abstract: The present paper is concerned with the Cauchy-Dirichlet problem for fractional (and non-fractional) nonlinear diffusion equations posed in bounded domains. Main results consist of well-posedness in an energy class with no sign restriction and convergence of such (possibly sign-changing) energy solutions to asymptotic profiles after a proper rescaling. They will be proved in a variational scheme only, without any use of semigroup theories nor classical quasilinear parabolic theories. Proofs are self-contained and performed in a totally unified fashion for both fractional and non-fractional cases as well as for both porous medium and fast diffusion cases.

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