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4.7.7: Applications of Linear Interpolation and Extrapolation

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Problem Solving with Linear Models

What if you've plotted some data points, with the x−coordinates of the points representing the number of years a teacher has been teaching at a school and the y−coordinates representing his salary? Suppose that you've found the line of best fit to be y=1500x+28,000. Could you use the line of best fit to predict how much his salary will be after he's taught for 12 years? How would you do it?

Linear Modeling

Previously we worked on writing equations and determining lines of best fit. When we fit a line to data using interpolation, extrapolation, or linear regression, it is called linear modeling.

A mathematical model is an equation that describes data. The data may be graphed in a scatter plot.

Let's solve the following problems using linear models:

• Dana heard something very interesting at school. Her teacher told her that if you divide the circumference of a circle by its diameter you always get the same number. She tested this statement by measuring the circumference and diameter of several circular objects. The following table shows her results. From this data, estimate the circumference of a circle whose diameter is 12 inches.

Begin by creating a scatter plot and drawing the line of best fit.

Find the equation of the line of best fit using points (0.25, 1.2) and (8, 25.5).

m=(25.5−12)/(8−0.25)=24.3/7.75=3.14 Slope

1.2=3.14(0.25)+b⇒b=0.42

y=3.14x+0.42 Equation

Diameter=12 inches⇒y=3.14(12)+0.42=38.1 inches

In this problem, the slope=3.14. This number should be very familiar to you—it is the number pi rounded to the hundredths place. Theoretically, the circumference of a circle divided by its diameter is always the same and it equals 3.14 or π.

• Using Dana's data from the previous problem, estimate the circumference of a circle whose diameter is 25 inches.

The equation y=3.14x+0.42 of the relationship between diameter and circumference from the previous problem applies here.

Diameter=25 inches⇒y=3.14(25)+0.42=78.92 inches

A circle with a diameter of 25 inches will have a circumference that is approximately 78.92 inches.

• Using Dana's data from the first problem, estimate the circumference of a circle whose diameter is 60 inches.

The equation y=3.14x+0.42 of the relationship between diameter and circumference from the first problem applies here.

Diameter=60 inches⇒y=3.14(60)+0.42=188.82 inches

A circle with a diameter of 60 inches will have a circumference that is approximately 188.82 inches.

Example 4.7.7.1

Earlier, you were told that you have plotted some data points with the x−coordinates of the points representing the number of years a teacher has been teaching at a school and the y−coordinates representing his salary. Suppose you've found the line of best fit to be y=1500x+28,000. Could you use the line of best fit to predict how much his salary will be after he's taught for 12 years?

The line of best fit is a linear model that represents the situation. Since you have a linear model, you can plug in 12 for x and solve for y. This would give you the salary of the teacher after he's taught for 12 years.

Example 4.7.7.2

A cylinder is filled with water to a height of 73 centimeters. The water is drained through a hole in the bottom of the cylinder and measurements are taken at two-second intervals. The table below shows the height of the water level in the cylinder at different times.

Find the water level at 15 seconds.

Begin by graphing the scatter plot. As you can see below, a straight line does not fit the majority of this data. Therefore, there is no line of best fit. Instead, use interpolation.

To find the value at 15 seconds, connect points (14, 21.9) and (16, 17.1) and find the equation of the straight line.

m=(17.1−21.9)/(16−14)=−4.8/2=−2.4

y=−2.4x+b⇒21.9=−2.4(14)+b⇒b=55.5

Equation y=−2.4x+55.5

Substitute x=15 and obtain y=−2.4(15)+55.5=19.5 cm.

• What is a mathematical model?
• What is linear modeling? What are the options to determine a linear model?
• Using the Water Level data, use interpolation to determine the height of the water at 17 seconds.

Use the Life Expectancy table below to answer the questions.

• Make a scatter plot of the data.
• Use a line of best fit to estimate the life expectancy of a person born in 1955.
• Use linear interpolation to estimate the life expectancy of a person born in 1955.
• Use a line of best fit to estimate the life expectancy of a person born in 1976.
• Use linear interpolation to estimate the life expectancy of a person born in 1976.
• Use a line of best fit to estimate the life expectancy of a person born in 2012.
• Use linear extrapolation to estimate the life expectancy of a person born in 2012.
• Which method gives better estimates for this data set? Why?

The table below lists the high temperature for the first day of each month in 2006 in San Diego, California (Weather Underground). Use this table to answer the questions.

• Draw a scatter plot of the data.
• Use a line of best fit to estimate the temperature in the middle of the 4th month (month 4.5).
• Use linear interpolation to estimate the temperature in the middle of the 4th month (month 4.5).
• Use a line of best fit to estimate the temperature for month 13 (January 2007).
• Use linear extrapolation to estimate the temperature for month 13 (January 2007).

Mixed Review

• Simplify 6t(−2+7t)−t(4+3t).
• Solve for y:119/8=−10/3(y+(7/5))−(5/2).
• Determine the final cost. Original cost of jacket: $45.00; 15% markup; and 8% sales tax.
• Write as a fraction: 0.096.
• Is this function an example of direct variation? g(t)=−35t+15. Explain your answer.
• Graph this data and connect the data points.
• What conclusions can you make regarding this data?
• There seems to be a large drop in school homicides between 1999 and 2001. What could have happened to cause such a large gap?
• Make a prediction about 2009 using this data.

Year 1993 1995 1997 1999 2001 2003 2005 2007

# 34 28 28 33 14 18 22 30

Review (Answers)

To see the Review answers, open this PDF file and look for section 5.12.

Additional Resources

Video: Solving Word Problems in Slope-Intercept Form - Example 1

Activity: Problem Solving with Linear Models Discussion Questions

Practice: Applications of Linear Interpolation and Extrapolation

Learn maths at home

Linear Programming: How to Find the Optimal Solution

Linear programming video lessons, how to do linear programming, linear programming with multiple solutions, linear programming with infinite solutions, linear programming with integer solutions, what is linear programming.

Linear programming is an algebraic method for finding an optimal value in a situation in which there are constraints. The process involves forming constraint equations, graphing the feasible region and substituting vertices into the objective function to find a minimum or maximum value.

• Constraint equations are found for each category. The amounts used in each category are compared to the total amount of each that is available or required.
• The feasible region is the shaded region between the constraint equations. All combinations of coordinates within this region satisfy the requirements of the problem.
• The vertices are the corners of the feasible region. These are the most extreme cases of the solutions found in the feasible region and therefore, the optimal solution is found at one or more of these vertices.
• The objective function is the equation which we want to maximise or minimise.

This process of testing the vertices in the objective function to find the optimal solution is known as the simplex method.

Linear programming is used in a variety of real life contexts such as:

• Manufacturing: Maximising profits by finding the optimal combination of products to manufacture.
• Engineering: Minimising wastage in a design whilst still meeting structural requirements.
• Business: Allocating resources and budgets to complete a project with minimal cost.
• Transportation: Minimising travel time when subject to limited resources or work force.
• Finance: Maximising profit or minimising risk of investments made in a portfolio.
• Scheduling: Minimising project completion time or constructing timetables.

How to Do Linear Programming to Find an Optimal Solution

• Form the constraint equations.
• Sketch the constraint equations.
• Find the vertices of the feasible region.
• Substitute the coordinates of the vertices into the objective function.
• The optimal solution is the vertex at which the maximum or minimum value is obtained.

Linear Programming: How to Find a Minimum

• The optimal solution is the vertex at which the minimum value is obtained.

A minimisation problem will involve searching for the feasible solution that minimises the objective function.

Typical minimisation problems involve the minimisation of cost, weight or time.

Here is an example of a minimisation problem in linear programming.

An athlete desires to have at least 60g of carbohydrates, 100g of protein and no more that 80g of fat each day.

He considers two different supplements to meet this requirement.

• Tins of supplement A cost $20 and contain 20g of carbohydrates, 25g of protein and 10g of fat.
• Tins of supplement B cost $30 and contain 20g of carbohydrates, 50g of protein and 20g of fat.

Find the combination of supplements that will meet his needs in the cheapest way.

In linear programming, it is helpful to write a paragraph of text as a table of ordered information as this allows the constraint equations to be found more easily.

Label each row of the table with the two items which we want to optimise.

In this example, that is supplement A and supplement B.

Label each column of the table with the categories that the two items contain.

In this example, each supplement contains carbohydrates, protein, fat and they have a cost. Therefore these are the names of each column.

The constraints are written by finding the total amount of each category that is available or required.

In this example, the athlete has a daily requirement for each of carbohydrate, protein and fat.

Since he requires at least 60g of carbohydrates, we write ≥ 60.

The amount of carbohydrates must be at least 60. This means it can be equal to 60 or greater but it cannot be less than 60.

Since he requires at least 100g of protein, we write ≥ 100. This means his total daily protein must be greater than or equal to 100.

However, since he must have no more than 80g of fat, he must have equal to 80g of fat or less than this. He cannot have more than 80g of fat. Therefore we write ≤ 80.

Step 1. Find the Constraint Equations

• Label the amount of the two items as x and y respectively.
• For each category, multiply the number each item contains by x or y respectively.
• Write the sum of these terms ≤ the total maximum limit or ≥ the total minimum requirement.

Here are some common meanings of the inequality signs.

In linear programming, when there is a minimum requirement for a particular resource we use the ≥ symbol.

When there is a fixed total maximum limit of a resource that is available, we use the ≤ symbol.

We will look at the constraint equation for each category individually.

Considering the carbohydrates constraint:

Supplement A (which we will label as x) contains 20g of carbohydrate.

So the total carbohydrate from A is equal to 20x.

Supplement B (which we will label as y) contains 20g of carbohydrate.

So the total carbohydrate from supplement B is equal to 20y.

The total carbohydrate from both supplements is therefore 20x + 20y.

In total, at least 60g of carbohydrate is required so we write ≥ 60.

Therefore, the constraint equation for carbohydrate is 20x + 20y ≥ 60.

This equation can be simplified easily by dividing all terms by 10 to obtain 2x + 2y ≥ 6.

It could also be simplified further by halving each term to obtain x + y ≥ 3 if needed.

The simplification of the constraint equations is not a necessary step.

Considering the protein constraint:

Each tin of supplement A contains 25g of protein, so the total protein from all of supplement A is 25x.

Each tin of supplement B contains 50g of protein, so the total protein from all of supplement B is 50y.

The total protein from both supplements is therefore 25x + 50y.

The athlete requires at least 100g of protein each day, so we write ≥100.

Therefore the constraint equation is 25x + 50y ≥ 100.

This can be simplified by dividing each term by 25 to obtain x + 2y ≥ 4.

Considering the fat constraint:

Each tin of supplement A contains 10g of fat, so the total fat from all of supplement A is 10x.

Each tin of supplement B contains 20g of fat, so the total fat from all of supplement B is 20y.

The athlete must have no more than 80g of fat in total so we write ≤ 80.

The inequality is now less than or equal to as the total fat must be less than or equal to 80.

The constraint is 10x + 20y ≤ 80.

This can be simplified by dividing each term by 10 to obtain x + 2y ≤ 8.

How to Find the Objective Function

The objective function is the equation that is to be maximised or minimised. It is of the form O = p x+ q y where x and y are the number of the two items being used and p and q are the constraints.

• If the cost is being minimised then p and q are the costs of x and y respectively.
• If the profit is being maximised then p and q are the profits from x and y respectively.
• If the weight is being minimised then p and q are the weights of x and y respectively.
• If the time is being minimised then p and q are the times of x and y respectively.
• If the volume is being maximised then p and q are the volumes of x and y respectively.

In this example, we are maximising cost. Therefore p and q are the costs of x and y respectively.

We write C = 20x + 30y.

Step 2. Sketch the Constraint Equations and Feasible Region

• Substitute x = 0 and solve for y to find the y-intercept.
• Substitute y = 0 and solve for x to find the x-intercept.
• Draw the line between the intercepts.
• Draw a horizontal line if the constraint is of the form y=a and a vertical line if it is of the form x=b.
• Shade below the line if y , left of the line if x .

A summary of where to shade for each given inequality is shown below.

If the constraint is of the form:

• x≤a then shade left of the line.
• x≥a then shade right of the line.
• y≤a then shade below the line.
• y≥a then shade above the line.
• x+y≤ a then shade below the line.
• x+y≥a then shade above the line.

Note that if the coefficients of x or y are negative, then the opposite will apply.

The first constraint to draw is 2x + 2y ≥ 6.

The x intercept is found by substituting y=0 into 2x + 2y = 6.

We obtain 2x = 6 and so, x = 3.

The y intercept is found by substituting x=0 into 2x + 2y = 6.

We obtain 2y = 6 and so, y = 3.

We connect the x and y intercepts with a straight line as shown below.

Since the equation contains a greater than or equal to sign (≥), we shade above the line.

The next constraint to draw is x + 2y ≥ 4.

The x intercept is found by substituting y=0 into x + 2y = 4.

We obtain x = 4 and so, x = 4.

The y intercept is found by substituting x=0 into x + 2y = 4.

We obtain 2y = 4 and so, y = 2.

The final constraint to draw is x + 2y ≤ 8.

The x intercept is found by substituting y=0 into x + 2y = 8.

We obtain x = 8.

The y intercept is found by substituting x=0 into x + 2y = 8.

We obtain 2y = 8 and so, y = 4.

Since the equation contains a greater than or equal to sign (≤), we shade below the line.

The feasible region is the region that satisfies all of the constraint equations. To graph the feasible region, shade above lines containing ‘≥’ and below lines containing ‘ ≤ ‘. If the coefficients of x or y are negative, the opposite applies.

In this example, we shade where the previously drawn lines are all pointing.

That is, above 2x+2y≥6. above x+2y≥4 and below x+2y≤8.

Step 3. Find the vertices of the feasible region.

The vertices of the feasible region are the corners of the shaded region. They are found wherever two or more constraint lines intersect. The optimal solution will be found at one or more of these vertices.

The vertices of this feasible region are:

(0,3) (0,4) (2,1) (4,0) and (8,0).

Step 4. Substitute the coordinates of the vertices into the objective function.

The vertices made up of a pair of coordinates (x, y) and are found at the corners of the feasible region. Substitute the values of x and y from these coordinates into the objective function to find the optimal solution.

We will do this ‘simplex process’ for each of the vertices.

The objective function is C = 20x + 30y.

Step 5. The optimal solution is the vertex at which the minimum value is obtained

In linear programming, the optimal solution is the maximum or minimum value of the objective function. This is always found at one of the vertices of the feasible region.

In this example, we wish to find the combination of supplements that meet the athlete’s needs at the cheapest cost.

The cheapest combination was the value which minimised the cost equation.

This was at the vertex (2, 1) with a cost of $70.

The vertex (2, 1) means that x=2 and y=1.

x is the number of supplement A that should be used.

y is the number of supplement B that should be used.

Therefore, the athlete should use 2 tins of supplement A and 1 tin of supplement B at a total cost of $70.

Linear Programming: How to Find a Maximum

• The optimal solution is the vertex at which the maxmimum value is obtained.

Here is an example of maximising an objective function using linear programming.

A restaurant produces two different breakfast options, small and large. Each day there are only 20 eggs and 12 tomatoes available to use in these breakfasts.

The small breakfast sells for $10 profit and contains 2 eggs and 1 tomato.

The large breakfast sells for $15 profit and contains 2 eggs and 2 tomatoes.

No more than 4 large breakfasts can be produced each day.

Find the number of each breakfast that should be made to maximise profit.

Step 1. Form the constraint equations.

We will let x be the number of small breakfasts produced each day.

We will let y be the number of large breakfasts produced each day.

Considering the eggs constraint:

The total eggs used in small breakfasts is 2x.

The total eggs used in large breakfasts is 2y.

There are only 20 eggs available therefore the total eggs used in all breakfasts must be less than or equal to 20.

2x + 2y ≤ 20.

We can simplify this by dividing each term by 2 to obtain x + y ≤ 10.

Considering the tomatoes constraint:

The number of tomatoes in the small breakfasts is just x.

The number of tomatoes in the large breakfasts is 2y.

There are only 12 tomatoes available so the total tomatoes must be less than or equal to 12.

x + 2y ≤ 12.

Considering that no more than 4 large breakfasts can be produced:

y is the number of large breakfasts. From this constraint, the number of large breakfasts must be less than or equal to 4.

Therefore y ≤ 4.

The objective function is the profit since we wish to maximise this.

The profit from the small breakfasts is 10x and the profit from the large breakfasts is 15y.

Therefore the objective function is P = 10x + 15y.

Considering the non-negative constraints:

We know that it is impossible to produce less than 0 of each type of breakfast, therefore the number of each type of breakfast must be greater than or equal to zero.

In linear programming, the non-negative constraints are x≥0 and y≥0. This means that the problem is confined to the positive quadrant involving only positive values of x and y.

They are included whenever the quantities being optimised cannot be negative, which is the case in the production or consumption found in real life problems.

Step 2. Graph the Constraint Equations

The non-negative constraints require us to only consider the top right quadrant with positive x and y values. The feasible region will be above the x-axis and to the right of the y-axis.

We have the three constraints x+y≤10, x+2y≤12 and y≤4.

Considering x+y≤10 :

The x-intercept is found by substituting y=0 into x+y=10. We obtain x=10.

The y-intercept is found by substituting x=0 into x+y=10. We obtain y=10.

We then connect these intercepts to find the line.

We shade below the line since the inequality in the constraint is ‘≤’.

Considering x+2y≤12 :

The x-intercept is found by substituting y=0 into x+2y=12. We obtain x=12.

The y-intercept is found by substituting x=0 into x+2y=12. We obtain 2y=12 and y = 6.

Considering y≤4 :

y=4 is a horizontal line at a height of y=4.

Step 3. Find the Vertices of the Feasible Region

The feasible region is found below all three of the given constraint equations.

This is the shaded region shown below.

The vertices of this region are: (0,4), (4, 4), (8, 2) and (10,0). There is also (0,0) but producing zero of each breakfast option will certainly not result in the optimal profit.

The objective function to be maximised is P = 10x + 15y.

Step 5. The optimal solution is the vertex at which the maximum value is obtained.

The optimal solution is produced from the vertex which maximises the profit.

This is found at (8, 2) where $110 profit is made.

This is the largest profit result produced.

Therefore 8 small breakfasts and 2 large breakfasts should be produced to maximise the profit and make $110.

How to Find Vertices in Linear Programming

The vertices can be observed by looking at the corners of the feasible region. Where the coordinates are not clear, vertices can be calculated accurately by solving the simultaneous equations formed by the two constraint equations that intersect at that particular point.

For example, consider the feasible region given by the constraints:

From the graph below it can be seen that the vertex lies at the coordinate (5, 3). This is where x=5 and y=3.

However, if we were not sure of this, we could find the coordinate of the vertex by solving the two constraint equations simultaneously.

To solve these equations simultaneously, we will multiply the equations so that there is a common term in both.

Multiplying equation (1) by 2, we obtain:

Multiplying equation (2) by 5, we obtain:

We write the equations above one another like so:

We subtract the top equation from the bottom equation to obtain:

Therefore, we obtain (5, 3) as the vertex.

Linear Programming with Multiple Optimal Solutions

Multiple optimal solutions occur if two or more vertices result in the same optimal solution. All points between these vertices will be optimal solutions too. Multiple solutions occur when the slope of the objective function is parallel to the constraint line connecting the multiple optimal solutions.

Here is an example of linear programming in which multiple optimal solutions occur.

A car manufacturer has access to 16 exhausts, 36 metal panels and 14 gearboxes.

He is able to manufacture 2 different models of car: Car model A and Car model B.

Car model A costs $20 000 and it requires 1 exhaust, 1 metal panel and 1 gearbox.

Car model B costs $40 000 and it requires 2 exhausts, 6 metal panels and 1 gearbox.

They wish to decide on the number of each car model to make to maximise their profits.

Since the total number of exhausts used must be less than or equal to 16, we obtain:

Since the total number of metal panels used must be less than or equal to 36, we obtain:

Since the total number of gearboxes used must be less than or equal to 14, we obtain:

The objective function to be maximised is the cost.

The constraints are drawn resulting in the feasible region shown below.

The vertices of the feasible region are: (0, 6), (6, 5), (12, 2) and (14, 0).

Substituting these into the objective function results in two maximum values.

Both (6, 5) and (12, 2) result in the objective function maximised at 320.

Since both (6, 5) and (12, 2) provide the optimal solution, we must also look at all solutions that lie on the line connecting these two points.

Since we are considering how many of each car model to manufacture, the solutions must be whole numbers rather than decimals.

So along with (6, 5) and (12, 2), we also have (8, 4) and (10, 3) as shown below.

Therefore the optimal solutions are (6, 5), (8, 4), (10, 3) and (12, 2).

All of these solutions will result in the same optimal objective function of 320.

This means that we should produce either of the following combinations to make a profit of $320 000:

6 of Car A, 5 of Car B

8 of Car A, 4 of Car B

10 of Car A, 3 of Car B

12 of Car A, 2 of Car B.

We will now look algebraically how multiple optimal solutions can be identified.

Multiple optimal solutions will be present when the objective function is parallel to a constraint of the feasible region. That is, the objective function will be a multiple of one of the constraint equations.

Therefore if the objective function is a multiple of one of the constraint equations, it will be parallel to it and multiple optimal solutions will lie on this line.

Therefore the optimised value of the objective function is 320.

The optimised value of the objective function is the value that the objective function is equal to so that when graphed, the objective function line lies on the boundary of the feasible region.

Infinite Solutions in Linear Programming

Linear programming problems can have infinite optimal solutions if two vertices result in the same optimised value of the objective function and the solutions can take non-integer values. The infinite solutions are found between the values of the two vertices.

Here is an example of a linear programming problem with infinite solutions.

A company has 18kg of gravel, 21kg of cement, 4kg of sand with which it can make two different concrete mixes.

Concrete A costs $2000 and it requires 3kg of gravel, 7kg of cement and 1kg of sand.

Concrete B costs $2000 and it requires 6kg of gravel, 3kg of cement and 1kg of sand.

The company can produce concrete in any measure quantity and wish to maximise their profits.

The constraint equations are given by:

The objective function is the cost of purchase, which we wish to maximise.

The feasible region is shown below all three of the constraint lines.

The vertices of the feasible region are: (0, 3), (2, 2), (2.25, 1.75) and (3,0).

Substituting these vertices into the objective function, we obtain two optimal answers of $8000 found at both (2,2) and (2.25, 1.75).

Since the concrete can be produced in any given quantity, the optimal solution does not need to take integer solutions and so, (2.25, 1.75) is a valid solution.

It simply means 2.25kg of concrete A and 1.75kg of concrete B.

Since there are two vertices which maximise the objective function, we have multiple solutions.

When two vertices optimise the objective function, they are both solutions and all points between these two points are also solutions.

Since the concrete does not need to take only integer values, it can take all decimal values between the points of (2,2) and (2.25, 1.75).

There are an infinite number of different decimal numbers this could be and therefore there are infinite solutions to this linear programming problem.

Whenever a linear problem has multiple solutions and can take non-integer values, there are an infinite number of solutions.

The green line below shows the range where the infinite solutions take place. All of these combinations of concrete result in a cost of purchase of $8000.

Integer Solutions to Linear Programming

If a linear programming solution can only take integer (whole number) values then also consider integer solutions inside the feasible region that are close to any decimal vertices and substitute these into the objective function as well.

Many linear programming problems deal with quantities that must take integer values.

Integers are whole numbers.

Only linear programming involving continuously measured quantities will permit non-integer solutions. Examples where non-integer solutions are allowed include problems involving weight, volume, capacity, area and length.

Here is an example in which only integer solutions are permitted.

A construction needs to be able to support at least 4 units of lateral load and at least 10 units of vertical load.

Steel beams weigh 50kg and support 1 unit of lateral load and 5 units of vertical load.

Wooden beams weigh 30kg and support 2 units of lateral load and 2 units of vertical load.

Find the optimal combination of beams that support the required loads but minimise the weight.

Since we cannot use fractional amounts of each beam, we can only use whole numbers in our solution. We can only take integer solutions.

The two constraint equations are shown graphed on the axes below.

We shade above both lines.

The vertices of the feasible region are (0, 5), (4,0) and (1.5, 1.25).

At (0,5) we the value of the objective function is 150 and at (4, 0) the value of the objective function is 200.

We do cannot use the third vertex of (1.5, 1.25) as it does not contain only whole numbers. We cannot use 1.5 or 1.25 of a beam in this problem.

Therefore we must look at whole number points that are inside the feasible region, near to this decimal vertex of (1.5, 1.25).

The closest whole number solutions are shown below with red coordinates.

We can look at (2, 2), (3, 1) and (1, 3).

We see that (2, 2) gives an objective function result of 160, (3, 1) results in 180 and (1, 3) results in 140.

Therefore the weight is minimised at (1, 3) with a weight of 140kg.

It is important to notice that the optimal solution in this example was found at (1, 3) which was not a vertex of the feasible region.

In all other instances, the optimal solution is always found at a vertex.

However, if one vertex is a decimal, it is possible that the optimal solution may be found inside the feasible region, nearby to it.

Linear Programming Without Graphing

Linear programming can be solved algebraically without graphing. Whilst graphing is often a faster process that assists the visualisation of the solution, it may not be practical for problems involving more constraints and linear programming can be solved in an algebraic manner or using a computer.

• Find the point of intersection for each of the possible different pairs of constraint equations.
• Substitute these points into the other constraint inequalities.
• Remove any points that fail the constraint inequalities.
• Substitute the remaining points into the objective function to find the optimal solution.

Here is an example of solving a linear programming constraint problem without graphing.

Step 1. Find the point of intersection for each of the possible different pairs of constraint equations

This results in a vertex at (2, 1).

This results in a vertex of (-2, 5).

There is no solution to this pair of equations.

We now continue by considering the solutions found by the following pairs of constraints.

The final list of vertices is therefore:

(0, 3), (0, 4), (2, 1), (4, 0), (8, 0).

These are then all substituted into the objective function to find the optimal result.

Linear Programming with 3 Variables

Linear programming problems with 3 variables can be solved graphically in 3 dimensions. 3D software is beneficial. Alternatively, it can be easier to solve linear programming with 3 or more variables computationally.

Here is an example in which a linear program problem involving 3 variables will be solved graphically.

We will see how linear programming constraint equations will be created in 3 variables: x, y and z.

A tour operator offers three different tours each day.

The City Tour requires 1 bus, 3 tour guides and 2 microphones. It makes $1000 profit.

The Country Tour requires 3 buses, 4 tour guides and 3 microphones. It makes $3000 profit.

The Scenic Tour requires 2 buses, 6 tour guides and 6 microphones. It makes $2500 in profit.

In total the tour operator has 20 buses, 50 tour guides and 42 microphones available to allocate.

The information is laid out as in the table above for clarity.

With linear programming with 3 variables 3 dimensions are considered in which each constraint equation forms a plane rather than a line.

The three constraint equations are depicted below as planes. The feasible region is bounded below each of these planes and the x, y and z axes.

The vertices of the feasible region can be found where each plane intersects the axes and also where the three planes intersect.

The vertices of this feasible region are where the three planes intersect and also where the planes intersect each axis.

The vertices of the feasible region are (16.67, 0, 0), (0, 0, 7), (0, 6.67, 0), (6, 2, 4), (0, 3, 5.5) and (8, 0, 4.33).

Substituting these values into the objective function, it is maximised at (0, 3, 5.5).

However since we cannot use decimal solutions (we cannot have 5.5 of a tour), we must search nearby to this vertex for other optimal solutions.

Instead the optimal solution is (1, 3, 5) which results in a profit of $22 500.

• School Guide
• Class 9 Syllabus
• Maths Notes Class 9
• Science Notes Class 9
• History Notes Class 9
• Geography Notes Class 9
• Political Science Notes Class 9
• NCERT Soln. Class 9 Maths
• RD Sharma Soln. Class 9
• Math Formulas Class 9
• Pair of Linear Equations in Two Variables
• Linear Equation in Two Variables
• Linear Equations in One Variable
• Graph of Linear Equations in Two Variables
• Practice Questions Class 10 Chapter 3 Pair of Linear Equations in Two Variables
• Introduction to Two-Variable Linear Equations in Straight Lines
• Graphical Methods of Solving Pair of Linear Equations in Two Variables
• Algebraic Methods of Solving Pair of Linear Equations in Two Variables
• Solve Linear Equations with Variable on both Sides

Practice Problem on Linear Equations in Two Variables

In this article, we will learn about one interesting topic which is covered in class 9 and class 10 mathematics. We will look at some formulas and problems of Linear equations in two variables.

Important Formulas on Linear Equations in Two Variables

• Linear equations in two variables are expressed in the form ax + by + c = 0, where a, b, and c are real numbers, and a and b are not both zero.
• The solution of the equation represents the values of x and x for which the equation holds true.
• if (a 1 /a 2 ≠ b 1 /b 2 ) then the equation has exactly one solution. The lines are intersecting lines.
• if (a 1 /a 2 = b 1 /b 2 = c 1 /c 2 ) then the equation has infinitely many solution. The lines are coincidental lines.
• if (a 1 /a 2 = b 1 /b 2 ≠ c 1 /c 2 ) then the equation has no solution. The lines are parallel lines.
• The slope of a line represented in the form y = mx + c is m, where m is the coefficient of x.

Practice Problems with Solutions

Q1. what are the coefficients of the equation 4x – 10y = 46.

To find the coefficients of the equation 4x – 10y = 46, we need to find the term which is multiplying the variable So, coefficient of x = 4 and coefficient of y = -10

Q2. What is the constant of the equation 4x – 10y = 46?

To find the constant of the equation 4x – 10y = 46, we need to find the term which is not multiplied with any variable So, constant of 4x – 10y – 46 = 0 is -46.

Q3. Is x = 3 and y = 10 a solution of the equation -14x + 12y = 30 ?

To check if a pair of values (x, y) is a solution of the equation -14x + 12y = 30, we need to verify that left hand side of equation should be equal to right hand side of equation i.e. L.H.S = R.H.S So, ⇒ -14x + 12y = 30 ⇒ -14 × 3 + 12 × 10 = 30 ⇒ -42 + 120 ≠ 30 So, LHS is not equal to RHS. So, x = 3 and y = 10 are not the solution of the equation -14x + 12y = 30

Q4. Is x = 3 and y = -10 a solution of the equation 10x + 3y = 0?

To check if a pair of values (x, y) is a solution of the equation 10x + 3y = 0, we need to verify that left hand side of equation should be equal to right hand side of equation i.e. L.H.S = R.H.S So, ⇒ 10x + 3y ⇒ 10 × 3 + 3 × (-10) ⇒ 30 – 30 ⇒ 0 So, LHS is equal to RHS. So, x = 3 and y = -10 are the solution of the equation 10x + 3y = 0

Q5. What’s the slope of the line 30x – 6y =3?

To find the slope of the line 30x – 6y = 3, follow these steps First, put the equation in the slope intercept form (y = mx + b) 6y = 30x – 3 y = 5x – 1/2 Now, check the coefficient of x Here the coefficient of x is 5 So, the slope of the line 30x – 6y = 3 is 5.

Q6. What’s the slope of the line -20x + 10y = 8?

To find the slope of the line -20x + 10y = 8, follow these steps First, put the equation in the slope intercept form (y = mx + b) 10y = 20x + 8 y = 2x + 4/5 Now, check the coefficient of x Here the coefficient of x is 2 So, the slope of the line 30x – 6y = 3 is 2.

Q7. Two Notebook and one pen cost Rs. 35 and 3 Notebook and four pen cost Rs. 65. Find the cost of Notebook and pen separately.

Let’s denote the cost of one notebook as N and the cost of one pen as P. 1. Two notebooks and one pen cost Rs. 35: 2N + 1P = 35 2. Three notebooks and four pens cost Rs. 65: 3N + 4P = 65 Let’s solve it using the elimination method: Multiplying the first equation by 4 and the second equation by 1 to eliminate P: 1. 4 * (2N + 1P) = 4 * 35 which gives 8N + 4P = 140 2. 1 * (3N + 4P) = 1 * 65 which gives 3N + 4P = 65 Now, subtracting the second equation from the first equation: (8N + 4P) – (3N + 4P) = 140 – 65 8N + 4P – 3N – 4P = 75 5N = 75 Dividing both sides by 5: N = 75/5 = 15 Now that we have found the cost of one notebook N = 15, we can substitute this value into one of the original equations to find the cost of one pen. From the first equation: 2N + 1P = 35 2(15) + 1P = 35 30 + P = 35 P = 35 – 30 P = 5 So, the cost of one notebook is Rs. 15 and the cost of one pen is Rs. 5.

Q8. Find the solution for the given pair of linear equations.

2x + 3y = 7, 4x – 6y = 10.

To find the number of solution , we check the ratio a 2 /a 1 = 4/2 = 2 and, b 2 /b 1 = -6/3 = -2 a 2 /a 1 ≠ b 2 /b 1 So, it have one solution. Now, to find the solution we have two equations 2x + 3y = 7 …..(i) 4x – 6y = 10 ……(ii) Multiply equation (i) by 2 4x + 6y = 14 …..(iii) Now add equation (ii) and (iii), we get 8x = 24 So, x = 3. Now the value of x in equation (i) 6 + 3y = 7 y = 1/3. So, x = 3 and y = 1/3.

Q9. Find the solution for the given pair of linear equations.

3x + 2y = 10, 6x + 4y = 20.

To find the number of solution , we check the ratio a 2 /a 1 = 6/3 = 2 and, b 2 /b 1 = 4/2 = 2 and, c 2 /c 1 = 20/10 = 2 Thus, a 2 /a 1 = b 2 /b 1 = c 2 /c 1 = 2 So, it have infinitely many solution. Now, to find the solution we have two equations 3x + 2y = 10 ….(i) 6x + 4y = 20 ….(ii) As, we observe that both lines are the same line. So, any point which fall on the line is the solution like, x = 2 and y = 2 x = 3 and y = 1/2 and many more.

Q10. Find the solution for the given pair of linear equations.

4x + 6y = 15.

To find the number of solution , we check the ratio a 2 /a 1 = 4/2 = 2 and, b 2 /b 1 = 6/3 = 2 and, c 2 /c 1 = 15/7 So, a 2 /a 1 = b 2 /b 1 ≠ c 2 /c 1 . So, it have no solution.

Problems on Linear Equations in Two Variables

P1. What are the coefficients of the equation 2x − 5y = 20?

P2. What is the constant of the equation 3x + 7y = −14?

P3. Is x=4 and y=2 a solution of the equation −5x + 3y = 7?

P4. Is x=−3 and y=5 a solution of the equation 8x − 2y = −34?

P5. What’s the slope of the line 6x − 9y = 12?

P6. What’s the slope of the line −4x + 8y = −16?

P7. Three apples and two oranges cost $8, and five apples and four oranges cost $18. Find the cost of an apple and an orange separately.

P8. Find the solution for the given pair of linear equations:

• 3x − 2y = 5
• 6x + 4y = 14

P9. Find the solution for the given pair of linear equations:

• 4x + 3y = 12
• 8x + 6y = 24

P10. Find the solution for the given pair of linear equations:

• 5x − 2y = 10
• 10x + 4y = 20

FAQs on Linear Equations in Two Variables

What are linear equations in two variables.

Linear equations in two variables are algebraic expressions involving two variables, typically denoted as x and y, with no variable raised to a power greater than 1.

How do you graph linear equations in two variables?

To graph a linear equation in two variables, ax + by = c, you can rearrange it into slope-intercept form, y=mx + b, where m is the slope and b is the y-intercept, then plot points using the slope and intercept.

What methods can be used to solve systems of linear equations in two variables?

Systems of linear equations in two variables can be solved using methods such as substitution, elimination, and graphing.

What is the significance of the slope in a linear equation?

The slope of a linear equation represents the rate of change of the dependent variable (y) with respect to the independent variable (x). It indicates the steepness of the line on the graph.

What does it mean when two linear equations in two variables are parallel?

If two linear equations in two variables have the same slope but different y-intercepts, they are parallel lines, and the system of equations has no solution. This indicates that the lines never intersect on the coordinate plane.

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Solve Systems of Nonlinear Equations

Learning objectives.

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

• Solve a system of nonlinear equations using graphing
• Solve a system of nonlinear equations using substitution
• Solve a system of nonlinear equations using elimination
• Use a system of nonlinear equations to solve applications

Solve a System of Nonlinear Equations Using Graphing

We learned how to solve systems of linear equations with two variables by graphing, substitution and elimination. We will be using these same methods as we look at nonlinear systems of equations with two equations and two variables. A system of nonlinear equations is a system where at least one of the equations is not linear.

For example each of the following systems is a system of nonlinear equations .

A system of nonlinear equations is a system where at least one of the equations is not linear.

Just as with systems of linear equations, a solution of a nonlinear system is an ordered pair that makes both equations true. In a nonlinear system, there may be more than one solution. We will see this as we solve a system of nonlinear equations by graphing.

When we solved systems of linear equations, the solution of the system was the point of intersection of the two lines. With systems of nonlinear equations, the graphs may be circles, parabolas or hyperbolas and there may be several points of intersection, and so several solutions. Once you identify the graphs, visualize the different ways the graphs could intersect and so how many solutions there might be.

To solve systems of nonlinear equations by graphing, we use basically the same steps as with systems of linear equations modified slightly for nonlinear equations. The steps are listed below for reference.

• Identify the graph of each equation. Sketch the possible options for intersection.
• Graph the first equation.
• Graph the second equation on the same rectangular coordinate system.
• Determine whether the graphs intersect.
• Identify the points of intersection.
• Check that each ordered pair is a solution to both original equations.

Solve a System of Nonlinear Equations Using Substitution

The graphing method works well when the points of intersection are integers and so easy to read off the graph. But more often it is difficult to read the coordinates of the points of intersection. The substitution method is an algebraic method that will work well in many situations. It works especially well when it is easy to solve one of the equations for one of the variables.

The substitution method is very similar to the substitution method that we used for systems of linear equations. The steps are listed below for reference.

• Solve one of the equations for either variable.
• Substitute the expression from Step 2 into the other equation.
• Solve the resulting equation.
• Substitute each solution in Step 4 into one of the original equations to find the other variable.
• Write each solution as an ordered pair.

No solution

So far, each system of nonlinear equations has had at least one solution. The next example will show another option.

Solve a System of Nonlinear Equations Using Elimination

When we studied systems of linear equations, we used the method of elimination to solve the system. We can also use elimination to solve systems of nonlinear equations. It works well when the equations have both variables squared. When using elimination, we try to make the coefficients of one variable to be opposites, so when we add the equations together, that variable is eliminated.

The elimination method is very similar to the elimination method that we used for systems of linear equations. The steps are listed for reference.

• Write both equations in standard form.
• Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites.
• Add the equations resulting from Step 3 to eliminate one variable.
• Solve for the remaining variable.
• Substitute each solution from Step 5 into one of the original equations. Then solve for the other variable.

There are also four options when we consider a circle and a hyperbola.

Use a System of Nonlinear Equations to Solve Applications

Systems of nonlinear equations can be used to model and solve many applications. We will look at an everyday geometric situation as our example.

The difference of the squares of two numbers is 15. The sum of the numbers is 5. Find the numbers.

Myra purchased a small 25” TV for her kitchen. The size of a TV is measured on the diagonal of the screen. The screen also has an area of 300 square inches. What are the length and width of the TV screen?

Edgar purchased a small 20” TV for his garage. The size of a TV is measured on the diagonal of the screen. The screen also has an area of 192 square inches. What are the length and width of the TV screen?

If the length is 12 inches, the width is 16 inches. If the length is 16 inches, the width is 12 inches.

The Harper family purchased a small microwave for their family room. The diagonal of the door measures 15 inches. The door also has an area of 108 square inches. What are the length and width of the microwave door?

If the length is 12 inches, the width is 9 inches. If the length is 9 inches, the width is 12 inches.

Access these online resources for additional instructions and practice with solving nonlinear equations.

• Nonlinear Systems of Equations
• Solve a System of Nonlinear Equations
• Solve a System of Nonlinear Equations by Elimination
• System of Nonlinear Equations – Area and Perimeter Application

Key Concepts

Section exercises, practice makes perfect.

In the following exercises, solve the system of equations by using graphing.

In the following exercises, solve the system of equations by using substitution.

In the following exercises, solve the system of equations by using elimination.

In the following exercises, solve the problem using a system of equations.

The sum of the squares of two numbers is 65. The difference of the number is 3. Find the numbers.

The sum of the squares of two numbers is 113. The difference of the number is 1. Find the numbers.

The difference of the squares of two numbers is 15. The difference of twice the square of the first number and the square of the second number is 30. Find the numbers.

The difference of the squares of two numbers is 20. The difference of the square of the first number and twice the square of the second number is 4. Find the numbers.

The perimeter of a rectangle is 32 inches and its area is 63 square inches. Find the length and width of the rectangle.

If the length is 11 cm, the width is 15 cm. If the length is 15 cm, the width is 11 cm.

Dion purchased a new microwave. The diagonal of the door measures 17 inches. The door also has an area of 120 square inches. What are the length and width of the microwave door?

Jules purchased a microwave for his kitchen. The diagonal of the front of the microwave measures 26 inches. The front also has an area of 240 square inches. What are the length and width of the microwave?

If the length is 10 inches, the width is 24 inches. If the length is 24 inches, the width is 10 inches.

Roman found a widescreen TV on sale, but isn’t sure if it will fit his entertainment center. The TV is 60”. The size of a TV is measured on the diagonal of the screen and a widescreen has a length that is larger than the width. The screen also has an area of 1728 square inches. His entertainment center has an insert for the TV with a length of 50 inches and width of 40 inches. What are the length and width of the TV screen and will it fit Roman’s entertainment center?

Donnette found a widescreen TV at a garage sale, but isn’t sure if it will fit her entertainment center. The TV is 50”. The size of a TV is measured on the diagonal of the screen and a widescreen has a length that is larger than the width. The screen also has an area of 1200 square inches. Her entertainment center has an insert for the TV with a length of 38 inches and width of 27 inches. What are the length and width of the TV screen and will it fit Donnette’s entertainment center?

The length is 40 inches and the width is 30 inches. The TV will not fit Donnette’s entertainment center.

Writing Exercises

In your own words, explain the advantages and disadvantages of solving a system of equations by graphing.

Explain in your own words how to solve a system of equations using substitution.

Answers will vary.

Explain in your own words how to solve a system of equations using elimination.

A circle and a parabola can intersect in ways that would result in 0, 1, 2, 3, or 4 solutions. Draw a sketch of each of the possibilities.

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

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Chapter Review Exercises

Distance and midpoint formulas; circles.

Use the Distance Formula

In the following exercises, find the distance between the points. Round to the nearest tenth if needed.

Use the Midpoint Formula

In the following exercises, find the midpoint of the line segments whose endpoints are given.

Write the Equation of a Circle in Standard Form

In the following exercises, write the standard form of the equation of the circle with the given information.

Graph a Circle

In the following exercises, ⓐ find the center and radius, then ⓑ graph each circle.

Graph Vertical Parabolas

In the following exercises, graph each equation by using its properties.

In the following exercises, ⓐ write the equation in standard form, then ⓑ use properties of the standard form to graph the equation.

Graph Horizontal Parabolas

Solve Applications with Parabolas

In the following exercises, create the equation of the parabolic arch formed in the foundation of the bridge shown. Give the answer in standard form.

Graph an Ellipse with Center at the Origin

In the following exercises, graph each ellipse.

Find the Equation of an Ellipse with Center at the Origin

In the following exercises, find the equation of the ellipse shown in the graph.

Graph an Ellipse with Center Not at the Origin

In the following exercises, ⓐ write the equation in standard form and ⓑ graph.

Solve Applications with Ellipses

In the following exercises, write the equation of the ellipse described.

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 10 AU and the furthest is approximately 90 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

In the following exercises, graph.

Identify the Graph of each Equation as a Circle, Parabola, Ellipse, or Hyperbola

In the following exercises, identify the type of graph.

ⓐ hyperbola ⓑ circle ⓒ parabola ⓓ ellipse

The sum of the squares of two numbers is 25. The difference of the numbers is 1. Find the numbers.

The difference of the squares of two numbers is 45. The difference of the square of the first number and twice the square of the second number is 9. Find the numbers.

The perimeter of a rectangle is 58 meters and its area is 210 square meters. Find the length and width of the rectangle.

If the length is 14 inches, the width is 15 inches. If the length is 15 inches, the width is 14 inches.

Colton purchased a larger microwave for his kitchen. The diagonal of the front of the microwave measures 34 inches. The front also has an area of 480 square inches. What are the length and width of the microwave?

Practice Test

In the following exercises, find the distance between the points and the midpoint of the line segment with the given endpoints. Round to the nearest tenth as needed.

Find the equation of the ellipse shown in the graph.

In the following exercises, ⓐ identify the type of graph of each equation as a circle, parabola, ellipse, or hyperbola, and ⓑ graph the equation.

ⓐ ellipse ⓑ

ⓐ hyperbola ⓑ

In the following exercises, ⓐ identify the type of graph of each equation as a circle, parabola, ellipse, or hyperbola, ⓑ write the equation in standard form, and ⓒ graph the equation.

A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately 20 AU and the furthest is approximately 70 AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.

For her birthday, Olive’s grandparents bought her a new widescreen TV. Before opening it she wants to make sure it will fit her entertainment center. The TV is 55”. The size of a TV is measured on the diagonal of the screen and a widescreen has a length that is larger than the width. The screen also has an area of 1452 square inches. Her entertainment center has an insert for the TV with a length of 50 inches and width of 40 inches. What are the length and width of the TV screen and will it fit Olive’s entertainment center?

The length is 44 inches and the width is 33 inches. The TV will fit Olive’s entertainment center.

Intermediate Algebra Copyright © 2017 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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● eqUations aND ineQUALIties Solving a value mixture problem using a linear equation 2/3 Jes __ At the city museum, child admission is $5.50 and adult admission is $9.50. On Saturday, twice as many adult tickets as child tickets were sold, for a total sales of $784.00. How many child tickets were sold that day? Number of child tickets: ×

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Ratios and proportions and how to solve them

Let's talk about ratios and proportions. When we talk about the speed of a car or an airplane we measure it in miles per hour. This is called a rate and is a type of ratio. A ratio is a way to compare two quantities by using division as in miles per hour where we compare miles and hours.

A ratio can be written in three different ways and all are read as "the ratio of x to y"

$$x\: to\: y$$

$$\frac{x}{y}$$

A proportion on the other hand is an equation that says that two ratios are equivalent. For instance if one package of cookie mix results in 20 cookies than that would be the same as to say that two packages will result in 40 cookies.

$$\frac{20}{1}=\frac{40}{2}$$

A proportion is read as "x is to y as z is to w"

$$\frac{x}{y}=\frac{z}{w} \: where\: y,w\neq 0$$

If one number in a proportion is unknown you can find that number by solving the proportion.

You know that to make 20 pancakes you have to use 2 eggs. How many eggs are needed to make 100 pancakes?

$$\frac{eggs}{pancakes}=\frac{eggs}{pancakes}\: \: or\: \: \frac{pancakes}{eggs}=\frac{pancakes}{eggs}$$

If we write the unknown number in the nominator then we can solve this as any other equation

$$\frac{x}{100}=\frac{2}{20}$$

Multiply both sides with 100

$${\color{green} {100\, \cdot }}\, \frac{x}{100}={\color{green} {100\, \cdot }}\, \frac{2}{20}$$

$$x=\frac{200}{20}$$

If the unknown number is in the denominator we can use another method that involves the cross product. The cross product is the product of the numerator of one of the ratios and the denominator of the second ratio. The cross products of a proportion is always equal

If we again use the example with the cookie mix used above

$$\frac{{\color{green} {20}}}{{\color{blue} {1}}}=\frac{{\color{blue} {40}}}{{\color{green} {2}}}$$

$${\color{blue} {1}}\cdot {\color{blue} {40}}={\color{green} {2}}\cdot {\color{green} {20}}=40$$

It is said that in a proportion if

If you look at a map it always tells you in one of the corners that 1 inch of the map correspond to a much bigger distance in reality. This is called a scaling. We often use scaling in order to depict various objects. Scaling involves recreating a model of the object and sharing its proportions, but where the size differs. One may scale up (enlarge) or scale down (reduce).  For example, the scale of 1:4 represents a fourth. Thus any measurement we see in the model would be 1/4 of the real measurement. If we wish to calculate the inverse, where we have a 20ft high wall and wish to reproduce it in the scale of 1:4, we simply calculate:

$$20\cdot 1:4=20\cdot \frac{1}{4}=5$$

In a scale model of 1:X where X is a constant, all measurements become 1/X - of the real measurement. The same mathematics applies when we wish to enlarge. Depicting something in the scale of 2:1 all measurements then become twice as large as in reality. We divide by 2 when we wish to find the actual measurement.

Video lesson

$$\frac{x}{x + 20} = \frac{24}{54}$$

• The coordinate plane
• Linear equations in the coordinate plane
• The slope of a linear function
• The slope-intercept form of a linear equation
• Writing linear equations using the slope-intercept form
• Writing linear equations using the point-slope form and the standard form
• Parallel and perpendicular lines
• Scatter plots and linear models
• Solving linear inequalities
• Solving compound inequalities
• Solving absolute value equations and inequalities
• Linear inequalities in two variables
• Graphing linear systems
• The substitution method for solving linear systems
• The elimination method for solving linear systems
• Systems of linear inequalities
• Properties of exponents
• Scientific notation
• Exponential growth functions
• Monomials and polynomials
• Special products of polynomials
• Polynomial equations in factored form
• Use graphing to solve quadratic equations
• Completing the square
• The quadratic formula
• The graph of a radical function
• Simplify radical expressions
• Radical equations
• The Pythagorean Theorem
• The distance and midpoint formulas
• Simplify rational expression
• Multiply rational expressions
• Division of polynomials
• Add and subtract rational expressions
• Solving rational equations
• Algebra 2 Overview
• Geometry Overview
• SAT Overview
• ACT Overview

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Watch CBS News

Teens come up with trigonometry proof for Pythagorean Theorem, a problem that stumped math world for centuries

By Bill Whitaker

May 5, 2024 / 7:00 PM EDT / CBS News

As the school year ends, many students will be only too happy to see math classes in their rearview mirrors. It may seem to some of us non-mathematicians that geometry and trigonometry were created by the Greeks as a form of torture, so imagine our amazement when we heard two high school seniors had proved a mathematical puzzle that was thought to be impossible for 2,000 years. 

We met Calcea Johnson and Ne'Kiya Jackson at their all-girls Catholic high school in New Orleans. We expected to find two mathematical prodigies.

Instead, we found at St. Mary's Academy , all students are told their possibilities are boundless.

Come Mardi Gras season, New Orleans is alive with colorful parades, replete with floats, and beads, and high school marching bands.

In a city where uniqueness is celebrated, St. Mary's stands out – with young African American women playing trombones and tubas, twirling batons and dancing - doing it all, which defines St. Mary's, students told us.

Junior Christina Blazio says the school instills in them they have the ability to accomplish anything. 

Christina Blazio: That is kinda a standard here. So we aim very high - like, our aim is excellence for all students. 

The private Catholic elementary and high school sits behind the Sisters of the Holy Family Convent in New Orleans East. The academy was started by an African American nun for young Black women just after the Civil War. The church still supports the school with the help of alumni.

In December 2022, seniors Ne'Kiya Jackson and Calcea Johnson were working on a school-wide math contest that came with a cash prize.

Ne'Kiya Jackson: I was motivated because there was a monetary incentive.

Calcea Johnson: 'Cause I was like, "$500 is a lot of money. So I-- I would like to at least try."

Both were staring down the thorny bonus question.

Bill Whitaker: So tell me, what was this bonus question?

Calcea Johnson: It was to create a new proof of the Pythagorean Theorem. And it kind of gave you a few guidelines on how would you start a proof.

The seniors were familiar with the Pythagorean Theorem, a fundamental principle of geometry. You may remember it from high school: a² + b² = c². In plain English, when you know the length of two sides of a right triangle, you can figure out the length of the third.

Both had studied geometry and some trigonometry, and both told us math was not easy. What no one told  them  was there had been more than 300 documented proofs of the Pythagorean Theorem using algebra and geometry, but for 2,000 years a proof using trigonometry was thought to be impossible, … and that was the bonus question facing them.

Bill Whitaker: When you looked at the question did you think, "Boy, this is hard"?

Ne'Kiya Jackson: Yeah. 

Bill Whitaker: What motivated you to say, "Well, I'm going to try this"?

Calcea Johnson: I think I was like, "I started something. I need to finish it." 

Bill Whitaker: So you just kept on going.

Calcea Johnson: Yeah.

For two months that winter, they spent almost all their free time working on the proof.

CeCe Johnson: She was like, "Mom, this is a little bit too much."

CeCe and Cal Johnson are Calcea's parents.

CeCe Johnson:   So then I started looking at what she really was doing. And it was pages and pages and pages of, like, over 20 or 30 pages for this one problem.

Cal Johnson: Yeah, the garbage can was full of papers, which she would, you know, work out the problems and-- if that didn't work she would ball it up, throw it in the trash. 

Bill Whitaker: Did you look at the problem? 

Neliska Jackson is Ne'Kiya's mother.

Neliska Jackson: Personally I did not. 'Cause most of the time I don't understand what she's doing (laughter).

Michelle Blouin Williams: What if we did this, what if I write this? Does this help? ax² plus ….

Their math teacher, Michelle Blouin Williams, initiated the math contest.

Bill Whitaker: And did you think anyone would solve it?

Michelle Blouin Williams: Well, I wasn't necessarily looking for a solve. So, no, I didn't—

Bill Whitaker: What were you looking for?

Michelle Blouin Williams: I was just looking for some ingenuity, you know—

Calcea and Ne'Kiya delivered on that! They tried to explain their groundbreaking work to 60 Minutes. Calcea's proof is appropriately titled the Waffle Cone.

Calcea Johnson: So to start the proof, we start with just a regular right triangle where the angle in the corner is 90°. And the two angles are alpha and beta.

Bill Whitaker: Uh-huh

Calcea Johnson: So then what we do next is we draw a second congruent, which means they're equal in size. But then we start creating similar but smaller right triangles going in a pattern like this. And then it continues for infinity. And eventually it creates this larger waffle cone shape.

Calcea Johnson: Am I going a little too—

Bill Whitaker: You've been beyond me since the beginning. (laughter) 

Bill Whitaker: So how did you figure out the proof?

Ne'Kiya Jackson: Okay. So you have a right triangle, 90° angle, alpha and beta.

Bill Whitaker: Then what did you do?

Ne'Kiya Jackson: Okay, I have a right triangle inside of the circle. And I have a perpendicular bisector at OP to divide the triangle to make that small right triangle. And that's basically what I used for the proof. That's the proof.

Bill Whitaker: That's what I call amazing.

Ne'Kiya Jackson: Well, thank you.

There had been one other documented proof of the theorem using trigonometry by mathematician Jason Zimba in 2009 – one in 2,000 years. Now it seems Ne'Kiya and Calcea have joined perhaps the most exclusive club in mathematics. 

Bill Whitaker: So you both independently came up with proof that only used trigonometry.

Ne'Kiya Jackson: Yes.

Bill Whitaker: So are you math geniuses?

Calcea Johnson: I think that's a stretch. 

Bill Whitaker: If not genius, you're really smart at math.

Ne'Kiya Jackson: Not at all. (laugh) 

To document Calcea and Ne'Kiya's work, math teachers at St. Mary's submitted their proofs to an American Mathematical Society conference in Atlanta in March 2023.

Ne'Kiya Jackson: Well, our teacher approached us and was like, "Hey, you might be able to actually present this," I was like, "Are you joking?" But she wasn't. So we went. I got up there. We presented and it went well, and it blew up.

Bill Whitaker: It blew up.

Calcea Johnson: Yeah. 

Ne'Kiya Jackson: It blew up.

Bill Whitaker: Yeah. What was the blowup like?

Calcea Johnson: Insane, unexpected, crazy, honestly.

It took millenia to prove, but just a minute for word of their accomplishment to go around the world. They got a write-up in South Korea and a shout-out from former first lady Michelle Obama, a commendation from the governor and keys to the city of New Orleans. 

Bill Whitaker: Why do you think so many people found what you did to be so impressive?

Ne'Kiya Jackson: Probably because we're African American, one. And we're also women. So I think-- oh, and our age. Of course our ages probably played a big part.

Bill Whitaker: So you think people were surprised that young African American women, could do such a thing?

Calcea Johnson: Yeah, definitely.

Ne'Kiya Jackson: I'd like to actually be celebrated for what it is. Like, it's a great mathematical achievement.

Achievement, that's a word you hear often around St. Mary's academy. Calcea and Ne'Kiya follow a long line of barrier-breaking graduates. 

The late queen of Creole cooking, Leah Chase , was an alum. so was the first African-American female New Orleans police chief, Michelle Woodfork …

And judge for the Fifth Circuit Court of Appeals, Dana Douglas. Math teacher Michelle Blouin Williams told us Calcea and Ne'Kiya are typical St. Mary's students.  

Bill Whitaker: They're not unicorns.

Michelle Blouin Williams: Oh, no no. If they are unicorns, then every single lady that has matriculated through this school is a beautiful, Black unicorn.

Pamela Rogers: You're good?

Pamela Rogers, St. Mary's president and interim principal, told us the students hear that message from the moment they walk in the door.

Pamela Rogers: We believe all students can succeed, all students can learn. It does not matter the environment that you live in. 

Bill Whitaker: So when word went out that two of your students had solved this almost impossible math problem, were they universally applauded?

Pamela Rogers: In this community, they were greatly applauded. Across the country, there were many naysayers.

Bill Whitaker: What were they saying?

Pamela Rogers: They were saying, "Oh, they could not have done it. African Americans don't have the brains to do it." Of course, we sheltered our girls from that. But we absolutely did not expect it to come in the volume that it came.  

Bill Whitaker: And after such a wonderful achievement.

Pamela Rogers: People-- have a vision of who can be successful. And-- to some people, it is not always an African American female. And to us, it's always an African American female.

Gloria Ladson-Billings: What we know is when teachers lay out some expectations that say, "You can do this," kids will work as hard as they can to do it.

Gloria Ladson-Billings, professor emeritus at the University of Wisconsin, has studied how best to teach African American students. She told us an encouraging teacher can change a life.

Bill Whitaker: And what's the difference, say, between having a teacher like that and a whole school dedicated to the excellence of these students?

Gloria Ladson-Billings: So a whole school is almost like being in Heaven. 

Bill Whitaker: What do you mean by that?

Gloria Ladson-Billings: Many of our young people have their ceilings lowered, that somewhere around fourth or fifth grade, their thoughts are, "I'm not going to be anything special." What I think is probably happening at St. Mary's is young women come in as, perhaps, ninth graders and are told, "Here's what we expect to happen. And here's how we're going to help you get there."

At St. Mary's, half the students get scholarships, subsidized by fundraising to defray the $8,000 a year tuition. Here, there's no test to get in, but expectations are high and rules are strict: no cellphones, modest skirts, hair must be its natural color.

Students Rayah Siddiq, Summer Forde, Carissa Washington, Tatum Williams and Christina Blazio told us they appreciate the rules and rigor.

Rayah Siddiq: Especially the standards that they set for us. They're very high. And I don't think that's ever going to change.

Bill Whitaker: So is there a heart, a philosophy, an essence to St. Mary's?

Summer Forde: The sisterhood—

Carissa Washington: Sisterhood.

Tatum Williams: Sisterhood.

Bill Whitaker: The sisterhood?

Voices: Yes.

Bill Whitaker: And you don't mean the nuns. You mean-- (laughter)

Christina Blazio: I mean, yeah. The community—

Bill Whitaker: So when you're here, there's just no question that you're going to go on to college.

Rayah Siddiq: College is all they talk about. (laughter) 

Pamela Rogers: … and Arizona State University (Cheering)

Principal Rogers announces to her 615 students the colleges where every senior has been accepted.

Bill Whitaker: So for 17 years, you've had a 100% graduation rate—

Pamela Rogers: Yes.

Bill Whitaker: --and a 100% college acceptance rate?

Pamela Rogers: That's correct.

Last year when Ne'Kiya and Calcea graduated, all their classmates went to college and got scholarships. Ne'Kiya got a full ride to the pharmacy school at Xavier University in New Orleans. Calcea, the class valedictorian, is studying environmental engineering at Louisiana State University.

Bill Whitaker: So wait a minute. Neither one of you is going to pursue a career in math?

Both: No. (laugh)

Calcea Johnson: I may take up a minor in math. But I don't want that to be my job job.

Ne'Kiya Jackson: Yeah. People might expect too much out of me if (laugh) I become a mathematician. (laugh)

But math is not completely in their rear-view mirrors. This spring they submitted their high school proofs for final peer review and publication … and are still working on further proofs of the Pythagorean Theorem. Since their first two …

Calcea Johnson: We found five. And then we found a general format that could potentially produce at least five additional proofs.

Bill Whitaker: And you're not math geniuses?

Bill Whitaker: I'm not buying it. (laughs)

Produced by Sara Kuzmarov. Associate producer, Mariah B. Campbell. Edited by Daniel J. Glucksman.

Bill Whitaker is an award-winning journalist and 60 Minutes correspondent who has covered major news stories, domestically and across the globe, for more than four decades with CBS News.

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