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Eureka Math Algebra 1 Module 3 Lesson 2 Answer Key

Engage ny eureka math algebra 1 module 3 lesson 2 answer key, eureka math algebra 1 module 3 lesson 2 example answer key.

Example 1. Consider Akelia’s sequence 5, 8, 11, 14, 17, …. a. If you believed in patterns, what might you say is the next number in the sequence? Answer: 20 (adding 3 each time)

b. Write a formula for Akelia’s sequence. Answer: A(n) = 5 + 3(n – 1)

c. Explain how each part of the formula relates to the sequence. Answer: To find each term in the sequence, you are adding 3 one less time than the term number. To get the 1st term, you add three zero times. To get the 2nd term, you add 3 one time. To get the 5th term, you add 3 four times.

d. Explain Johnny’s formula. Answer: His formula is saying that to find any term in the sequence, just add 3 to the term before it. For example, to find the 12th term, add 3 to the 11th term: A(12) = A(11) + 3. To find the 50th term, add 3 to the 49th term. To find the (n + 1)th term, add 3 to the nth term. It is critical that the value of the very first term be specified; we need it to get started finding the values of all the other terms.

Example 2. Consider a sequence given by the formula a n = a (n-1) -5, where a 1 = 12 and n ≥ 2. a. List the first five terms of the sequence. Answer: 12, 7, 2, -3, -8

b. Write an explicit formula. Answer: a n = 12-5(n-1) for n ≥ 1

c. Find a_6 and a_100 of the sequence. Answer: a 6 = -13 a 100 = -483

Eureka Math Algebra 1 Module 3 Lesson 2 Exercise Answer Key

Exercises 1–2

Exercise 1. Akelia, in a playful mood, asked Johnny: “What would happen if we change the ‘ + ’ sign in your formula to a ‘-’ sign? To a ‘×’ sign? To a ‘ ÷ ’ sign?” a. What sequence does A(n + 1) = A(n)-3 for n ≥ 1 and A(1) = 5 generate? Answer: Answer: 5, 2,-1, -4, …

b. What sequence does A(n + 1) = A(n) ⋅ 3 for n ≥ 1 and A(1) = 5 generate? Answer: 5, 15, 45, 135, …

c. What sequence does A(n + 1) = A(n) ÷ 3 for n ≥ 1 and A(1) = 5 generate? Answer: 5, \(\frac{5}{3}\), \(\frac{5}{9}\), \(\frac{5}{27}\), …

Exercise 2. Ben made up a recursive formula and used it to generate a sequence. He used B(n) to stand for the nth term of his recursive sequence. a. What does B(3) mean? Answer: It is the third term of Ben’s sequence.

b. What does B(m) mean? Answer: It is the m th term of Ben’s sequence.

c. If B(n + 1) = 33 and B(n) = 28, write a possible recursive formula involving B(n + 1) and B(n) that would generate 28 and 33 in the sequence. Answer: B(n) = B(n-1) + 5 (Note that this is not the only possible answer; it assumes the sequence is arithmetic and is probably the most obvious response students will give. If the sequence were geometric, the answer could be written as B(n + 1) = (\(\frac{33}{28}\))B(n).)

d. What does 2B(7) + 6 mean? Answer: It is 2 times the 7th term of Ben’s sequence plus 6.

e. What does B(n) + B(m) mean? Answer: It is the sum of the nth term of Ben’s sequence plus the m th term of Ben’s sequence.

f. Would it necessarily be the same as B(n + m)? Answer: No, adding two terms of a sequence is not the same as adding two of the term numbers and then finding that term of a sequence. Consider, for example, the sequence 1, 3, 5, 7, 9, 11, 13, …. Adding the 2nd and 3rd terms does not give you the 5th term.

g. What does B(17)-B(16) mean? Answer: It is the 17th term of Ben’s sequence minus the 16th term of Ben’s sequence.

Exercises 3–6

Exercise 3. One of the most famous sequences is the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, …. f(n + 1) = f(n) + f(n – 1), where f(1) = 1, f(2) = 1, and n ≥ 2 How is each term of the sequence generated? Answer: By adding the two preceding terms

Exercise 4. Each sequence below gives an explicit formula. Write the first five terms of each sequence. Then, write a recursive formula for the sequence. a. a n = 2n + 10 for n ≥ 1 Answer: 12, 14, 16, 18, 20 a n + 1 = a n + 2, where a 1 = 12 and n ≥ 1

b. a_n = (\(\frac{1}{2}\)) (n-1) for n ≥ 1 Answer: 1, \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\) a n + 1 = a n ÷ 2, where a 1 = 1 and n ≥ 1

Exercise 5. For each sequence, write either an explicit or a recursive formula. a. 1, -1, 1, -1, 1, -1, … Answer: a (n + 1) -a n , where a 1 = 1 and n≥1 or f(n) = (-1) (n + 1) , where n ≥ 1

b. \(\frac{1}{2}\), \(\frac{2}{3}\), \(\frac{3}{4}\), \(\frac{4}{5}\), … Answer: f(n) = \(\frac{n}{n + 1}\) and n ≥ 1

Exercise 6. Lou opens a bank account. The deal he makes with his mother is that if he doubles the amount that was in the account at the beginning of each month by the end of the month, she will add an additional $5 to the account at the end of the month. a. Let A(n) represent the amount in the account at the beginning of the nth month. Assume that he does, in fact, double the amount every month. Write a recursive formula for the amount of money in his account at the beginning of the (n + 1)th month. Answer: A(n + 1) = 2A(n) + 5, where n ≥ 1 and A(1) is the initial amount

b. What is the least amount he could start with in order to have $300 by the beginning of the third month? Answer: A(3) = 2 ∙ A(2) + 5 A(3) = 2 ∙ [2 ∙ A(1) + 5] + 5 300 ≤ 2 ∙ [2 ∙ A(1) + 5] + 5 300 ≤ 4 ∙ A(1) + 15 71.25 ≤ A(1) The least amount he could start with in order to have $300 by the beginning of the third month is $71.25.

Eureka Math Algebra 1 Module 3 Lesson 2 Problem Set Answer Key

For Problems 1–4, list the first five terms of each sequence.

Question 1. a n + 1 = a n + 6, where a 1 = 11 for n ≥ 1 Answer: 11, 17, 23, 29, 35

Question 2. a n = a n-1 ÷ 2, where a 1 = 50 for n ≥ 2 Answer: 50, 25, 12.5, 6.25, 3.125

Question 3. f(n + 1) = -2f(n) + 8 and f(1) = 1 for n ≥ 1 Answer: 1, 6, -4, 16, -24

Question 4. f(n) = f(n-1) + n and f(1) = 4 for n ≥ 2 Answer: 4, 6, 9, 13, 18

For Problems 5–10, write a recursive formula for each sequence given or described below.

Question 5. It follows a plus one pattern: 8, 9, 10, 11, 12, Answer: f(n + 1) = f(n) + 1, where f(1) = 8 and n ≥ 1

Question 6. It follows a times 10 pattern: 4, 40, 400, 4000, …. Answer: f(n + 1) = 10f(n), where f(1) = 4 and n ≥ 1

Question 7. It has an explicit formula of f(n) = -3n + 2 for n ≥ 1. Answer: (n + 1) = f(n)-3, where f(1) = -1 and n ≥ 1

Question 8. It has an explicit formula of f(n) = -1(12) (n-1) for n ≥ 1. Answer: f(n + 1) = 12f(n), where f(1) = -1 for n ≥ 1

Question 9. Doug accepts a job where his starting salary is $30,000 per year, and each year he receives a raise of $3,000. Answer: D (n + 1) = D n + 3000, where D 1 = 30000 and n ≥ 1

Question 10. A bacteria culture has an initial population of 10 bacteria, and each hour the population triples in size. Answer: B (n + 1) = 3B n , where B 1 = 10 and n ≥ 1

Eureka Math Algebra 1 Module 3 Lesson 2 Exit Ticket Answer Key

Question 1. Consider the sequence following a minus 8 pattern: 9, 1, -7, -15, …. a. Write an explicit formula for the sequence. Answer: f(n) = 9-8(n-1) for n ≥ 1

b. Write a recursive formula for the sequence. Answer: f(n + 1) = f(n)-8 and f(1) = 9 for n ≥ 1

c. Find the 38th term of the sequence. Answer: f(38) = 9-8(37) = -287

Question 2. Consider the sequence given by the formula a(n + 1) = 5a(n) and a(1) = 2 for n ≥ 1. a. Explain what the formula means. Answer: The first term of the sequence is 2. Each subsequent term of the sequence is found by multiplying the previous term by 5.

b. List the first five terms of the sequence. Answer: 2, 10, 50, 250, 1250

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2.1 The Rectangular Coordinate Systems and Graphs

x -intercept is ( 4 , 0 ) ; ( 4 , 0 ) ; y- intercept is ( 0 , 3 ) . ( 0 , 3 ) .

125 = 5 5 125 = 5 5

( − 5 , 5 2 ) ( − 5 , 5 2 )

2.2 Linear Equations in One Variable

x = −5 x = −5

x = −3 x = −3

x = 10 3 x = 10 3

x = 1 x = 1

x = − 7 17 . x = − 7 17 . Excluded values are x = − 1 2 x = − 1 2 and x = − 1 3 . x = − 1 3 .

x = 1 3 x = 1 3

m = − 2 3 m = − 2 3

y = 4 x −3 y = 4 x −3

x + 3 y = 2 x + 3 y = 2

Horizontal line: y = 2 y = 2

Parallel lines: equations are written in slope-intercept form.

y = 5 x + 3 y = 5 x + 3

2.3 Models and Applications

C = 2.5 x + 3 , 650 C = 2.5 x + 3 , 650

L = 37 L = 37 cm, W = 18 W = 18 cm

2.4 Complex Numbers

−24 = 0 + 2 i 6 −24 = 0 + 2 i 6

( 3 −4 i ) − ( 2 + 5 i ) = 1 −9 i ( 3 −4 i ) − ( 2 + 5 i ) = 1 −9 i

5 2 − i 5 2 − i

18 + i 18 + i

−3 −4 i −3 −4 i

2.5 Quadratic Equations

( x − 6 ) ( x + 1 ) = 0 ; x = 6 , x = − 1 ( x − 6 ) ( x + 1 ) = 0 ; x = 6 , x = − 1

( x −7 ) ( x + 3 ) = 0 , ( x −7 ) ( x + 3 ) = 0 , x = 7 , x = 7 , x = −3. x = −3.

( x + 5 ) ( x −5 ) = 0 , ( x + 5 ) ( x −5 ) = 0 , x = −5 , x = −5 , x = 5. x = 5.

( 3 x + 2 ) ( 4 x + 1 ) = 0 , ( 3 x + 2 ) ( 4 x + 1 ) = 0 , x = − 2 3 , x = − 2 3 , x = − 1 4 x = − 1 4

x = 0 , x = −10 , x = −1 x = 0 , x = −10 , x = −1

x = 4 ± 5 x = 4 ± 5

x = 3 ± 22 x = 3 ± 22

x = − 2 3 , x = − 2 3 , x = 1 3 x = 1 3

2.6 Other Types of Equations

{ −1 } { −1 }

0 , 0 , 1 2 , 1 2 , − 1 2 − 1 2

1 ; 1 ; extraneous solution − 2 9 − 2 9

−2 ; −2 ; extraneous solution −1 −1

−1 , −1 , 3 2 3 2

−3 , 3 , − i , i −3 , 3 , − i , i

2 , 12 2 , 12

−1 , −1 , 0 0 is not a solution.

2.7 Linear Inequalities and Absolute Value Inequalities

[ −3 , 5 ] [ −3 , 5 ]

( − ∞ , −2 ) ∪ [ 3 , ∞ ) ( − ∞ , −2 ) ∪ [ 3 , ∞ )

x < 1 x < 1

x ≥ −5 x ≥ −5

( 2 , ∞ ) ( 2 , ∞ )

[ − 3 14 , ∞ ) [ − 3 14 , ∞ )

6 < x ≤ 9 ​ or ( 6 , 9 ] 6 < x ≤ 9 ​ or ( 6 , 9 ]

( − 1 8 , 1 2 ) ( − 1 8 , 1 2 )

| x −2 | ≤ 3 | x −2 | ≤ 3

k ≤ 1 k ≤ 1 or k ≥ 7 ; k ≥ 7 ; in interval notation, this would be ( − ∞ , 1 ] ∪ [ 7 , ∞ ) . ( − ∞ , 1 ] ∪ [ 7 , ∞ ) .

2.1 Section Exercises

Answers may vary. Yes. It is possible for a point to be on the x -axis or on the y -axis and therefore is considered to NOT be in one of the quadrants.

The y -intercept is the point where the graph crosses the y -axis.

The x- intercept is ( 2 , 0 ) ( 2 , 0 ) and the y -intercept is ( 0 , 6 ) . ( 0 , 6 ) .

The x- intercept is ( 2 , 0 ) ( 2 , 0 ) and the y -intercept is ( 0 , −3 ) . ( 0 , −3 ) .

The x- intercept is ( 3 , 0 ) ( 3 , 0 ) and the y -intercept is ( 0 , 9 8 ) . ( 0 , 9 8 ) .

y = 4 − 2 x y = 4 − 2 x

y = 5 − 2 x 3 y = 5 − 2 x 3

y = 2 x − 4 5 y = 2 x − 4 5

d = 74 d = 74

d = 36 = 6 d = 36 = 6

d ≈ 62.97 d ≈ 62.97

( 3 , − 3 2 ) ( 3 , − 3 2 )

( 2 , −1 ) ( 2 , −1 )

( 0 , 0 ) ( 0 , 0 )

y = 0 y = 0

not collinear

A: ( −3 , 2 ) , B: ( 1 , 3 ) , C: ( 4 , 0 ) A: ( −3 , 2 ) , B: ( 1 , 3 ) , C: ( 4 , 0 )

d = 8.246 d = 8.246

d = 5 d = 5

( −3 , 4 ) ( −3 , 4 )

x = 0          y = −2 x = 0          y = −2

x = 0.75 y = 0 x = 0.75 y = 0

x = − 1.667 y = 0 x = − 1.667 y = 0

15 − 11.2 = 3.8 mi 15 − 11.2 = 3.8 mi shorter

6 .0 42 6 .0 42

Midpoint of each diagonal is the same point ( 2 , –2 ) ( 2 , –2 ) . Note this is a characteristic of rectangles, but not other quadrilaterals.

2.2 Section Exercises

It means they have the same slope.

The exponent of the x x variable is 1. It is called a first-degree equation.

If we insert either value into the equation, they make an expression in the equation undefined (zero in the denominator).

x = 2 x = 2

x = 2 7 x = 2 7

x = 6 x = 6

x = 3 x = 3

x = −14 x = −14

x ≠ −4 ; x ≠ −4 ; x = −3 x = −3

x ≠ 1 ; x ≠ 1 ; when we solve this we get x = 1 , x = 1 , which is excluded, therefore NO solution

x ≠ 0 ; x ≠ 0 ; x = − 5 2 x = − 5 2

y = − 4 5 x + 14 5 y = − 4 5 x + 14 5

y = − 3 4 x + 2 y = − 3 4 x + 2

y = 1 2 x + 5 2 y = 1 2 x + 5 2

y = −3 x − 5 y = −3 x − 5

y = 7 y = 7

y = −4 y = −4

8 x + 5 y = 7 8 x + 5 y = 7

Perpendicular

m = − 9 7 m = − 9 7

m = 3 2 m = 3 2

m 1 = − 1 3 ,   m 2 = 3 ;   Perpendicular . m 1 = − 1 3 ,   m 2 = 3 ;   Perpendicular .

y = 0.245 x − 45.662. y = 0.245 x − 45.662. Answers may vary. y min = −50 , y max = −40 y min = −50 , y max = −40

y = − 2.333 x + 6.667. y = − 2.333 x + 6.667. Answers may vary. y min = −10 ,   y max = 10 y min = −10 ,   y max = 10

y = − A B x + C B y = − A B x + C B

The slope for  ( −1 , 1 ) to  ( 0 , 4 ) is  3. The slope for  ( −1 , 1 ) to  ( 2 , 0 ) is  − 1 3 . The slope for  ( 2 , 0 ) to  ( 3 , 3 ) is  3. The slope for  ( 0 , 4 ) to  ( 3 , 3 ) is  − 1 3 . The slope for  ( −1 , 1 ) to  ( 0 , 4 ) is  3. The slope for  ( −1 , 1 ) to  ( 2 , 0 ) is  − 1 3 . The slope for  ( 2 , 0 ) to  ( 3 , 3 ) is  3. The slope for  ( 0 , 4 ) to  ( 3 , 3 ) is  − 1 3 .

Yes they are perpendicular.

2.3 Section Exercises

Answers may vary. Possible answers: We should define in words what our variable is representing. We should declare the variable. A heading.

2 , 000 − x 2 , 000 − x

v + 10 v + 10

Ann: 23 ; 23 ; Beth: 46 46

20 + 0.05 m 20 + 0.05 m

90 + 40 P 90 + 40 P

50 , 000 − x 50 , 000 − x

She traveled for 2 h at 20 mi/h, or 40 miles.

$5,000 at 8% and $15,000 at 12%

B = 100 + .05 x B = 100 + .05 x

R = 9 R = 9

r = 4 5 r = 4 5 or 0.8

W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14 W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14

f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21 f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21

m = − 5 4 m = − 5 4

h = 2 A b 1 + b 2 h = 2 A b 1 + b 2

length = 360 ft; width = 160 ft

A = 88 in . 2 A = 88 in . 2

h = V π r 2 h = V π r 2

r = V π h r = V π h

C = 12 π C = 12 π

2.4 Section Exercises

Add the real parts together and the imaginary parts together.

Possible answer: i i times i i equals -1, which is not imaginary.

−8 + 2 i −8 + 2 i

14 + 7 i 14 + 7 i

− 23 29 + 15 29 i − 23 29 + 15 29 i

8 − i 8 − i

−11 + 4 i −11 + 4 i

2 −5 i 2 −5 i

6 + 15 i 6 + 15 i

−16 + 32 i −16 + 32 i

−4 −7 i −4 −7 i

2 − 2 3 i 2 − 2 3 i

4 − 6 i 4 − 6 i

2 5 + 11 5 i 2 5 + 11 5 i

1 + i 3 1 + i 3

( 3 2 + 1 2 i ) 6 = −1 ( 3 2 + 1 2 i ) 6 = −1

5 −5 i 5 −5 i

9 2 − 9 2 i 9 2 − 9 2 i

2.5 Section Exercises

It is a second-degree equation (the highest variable exponent is 2).

We want to take advantage of the zero property of multiplication in the fact that if a ⋅ b = 0 a ⋅ b = 0 then it must follow that each factor separately offers a solution to the product being zero: a = 0 o r b = 0. a = 0 o r b = 0.

One, when no linear term is present (no x term), such as x 2 = 16. x 2 = 16. Two, when the equation is already in the form ( a x + b ) 2 = d . ( a x + b ) 2 = d .

x = 6 , x = 6 , x = 3 x = 3

x = − 5 2 , x = − 5 2 , x = − 1 3 x = − 1 3

x = 5 , x = 5 , x = −5 x = −5

x = − 3 2 , x = − 3 2 , x = 3 2 x = 3 2

x = −2 , 3 x = −2 , 3

x = 0 , x = 0 , x = − 3 7 x = − 3 7

x = −6 , x = −6 , x = 6 x = 6

x = 6 , x = 6 , x = −4 x = −4

x = 1 , x = 1 , x = −2 x = −2

x = −2 , x = −2 , x = 11 x = 11

z = 2 3 , z = 2 3 , z = − 1 2 z = − 1 2

x = 3 ± 17 4 x = 3 ± 17 4

One rational

Two real; rational

x = − 1 ± 17 2 x = − 1 ± 17 2

x = 5 ± 13 6 x = 5 ± 13 6

x = − 1 ± 17 8 x = − 1 ± 17 8

x ≈ 0.131 x ≈ 0.131 and x ≈ 2.535 x ≈ 2.535

x ≈ − 6.7 x ≈ − 6.7 and x ≈ 1.7 x ≈ 1.7

a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a

x ( x + 10 ) = 119 ; x ( x + 10 ) = 119 ; 7 ft. and 17 ft.

maximum at x = 70 x = 70

The quadratic equation would be ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. The two values of x x are 20 and 60.

2.6 Section Exercises

This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.

He or she is probably trying to enter negative 9, but taking the square root of −9 −9 is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in −27. −27.

A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.

x = 81 x = 81

x = 17 x = 17

x = 8 ,     x = 27 x = 8 ,     x = 27

x = −2 , 1 , −1 x = −2 , 1 , −1

y = 0 ,     3 2 ,     − 3 2 y = 0 ,     3 2 ,     − 3 2

m = 1 , −1 m = 1 , −1

x = 2 5 , ±3 i x = 2 5 , ±3 i

x = 32 x = 32

t = 44 3 t = 44 3

x = −2 x = −2

x = 4 , −4 3 x = 4 , −4 3

x = − 5 4 , 7 4 x = − 5 4 , 7 4

x = 3 , −2 x = 3 , −2

x = 1 , −1 , 3 , -3 x = 1 , −1 , 3 , -3

x = 2 , −2 x = 2 , −2

x = 1 , 5 x = 1 , 5

x ≥ 0 x ≥ 0

x = 4 , 6 , −6 , −8 x = 4 , 6 , −6 , −8

2.7 Section Exercises

When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.

( − ∞ , ∞ ) ( − ∞ , ∞ )

We start by finding the x -intercept, or where the function = 0. Once we have that point, which is ( 3 , 0 ) , ( 3 , 0 ) , we graph to the right the straight line graph y = x −3 , y = x −3 , and then when we draw it to the left we plot positive y values, taking the absolute value of them.

( − ∞ , 3 4 ] ( − ∞ , 3 4 ]

[ − 13 2 , ∞ ) [ − 13 2 , ∞ )

( − ∞ , 3 ) ( − ∞ , 3 )

( − ∞ , − 37 3 ] ( − ∞ , − 37 3 ]

All real numbers ( − ∞ , ∞ ) ( − ∞ , ∞ )

( − ∞ , − 10 3 ) ∪ ( 4 , ∞ ) ( − ∞ , − 10 3 ) ∪ ( 4 , ∞ )

( − ∞ , −4 ] ∪ [ 8 , + ∞ ) ( − ∞ , −4 ] ∪ [ 8 , + ∞ )

No solution

( −5 , 11 ) ( −5 , 11 )

[ 6 , 12 ] [ 6 , 12 ]

[ −10 , 12 ] [ −10 , 12 ]

x > − 6 and x > − 2 Take the intersection of two sets . x > − 2 ,   ( − 2 , + ∞ ) x > − 6 and x > − 2 Take the intersection of two sets . x > − 2 ,   ( − 2 , + ∞ )

x < − 3   or   x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ ​ ​ [ 1 , ∞ ) x < − 3   or   x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ ​ ​ [ 1 , ∞ )

( − ∞ , −1 ) ∪ ( 3 , ∞ ) ( − ∞ , −1 ) ∪ ( 3 , ∞ )

[ −11 , −3 ] [ −11 , −3 ]

It is never less than zero. No solution.

Where the blue line is above the orange line; point of intersection is x = − 3. x = − 3.

( − ∞ , −3 ) ( − ∞ , −3 )

Where the blue line is above the orange line; always. All real numbers.

( − ∞ , − ∞ ) ( − ∞ , − ∞ )

( −1 , 3 ) ( −1 , 3 )

( − ∞ , 4 ) ( − ∞ , 4 )

{ x | x < 6 } { x | x < 6 }

{ x | −3 ≤ x < 5 } { x | −3 ≤ x < 5 }

( −2 , 1 ] ( −2 , 1 ]

( − ∞ , 4 ] ( − ∞ , 4 ]

Where the blue is below the orange; always. All real numbers. ( − ∞ , + ∞ ) . ( − ∞ , + ∞ ) .

Where the blue is below the orange; ( 1 , 7 ) . ( 1 , 7 ) .

x = 2 , − 4 5 x = 2 , − 4 5

( −7 , 5 ] ( −7 , 5 ]

80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400 80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400

[ 1 , 600 , 2 , 400 ] [ 1 , 600 , 2 , 400 ]

Review Exercises

x -intercept: ( 3 , 0 ) ; ( 3 , 0 ) ; y -intercept: ( 0 , −4 ) ( 0 , −4 )

y = 5 3 x + 4 y = 5 3 x + 4

72 = 6 2 72 = 6 2

620.097 620.097

midpoint is ( 2 , 23 2 ) ( 2 , 23 2 )

x = 4 x = 4

x = 12 7 x = 12 7

y = 1 6 x + 4 3 y = 1 6 x + 4 3

y = 2 3 x + 6 y = 2 3 x + 6

females 17, males 56

x = − 3 4 ± i 47 4 x = − 3 4 ± i 47 4

horizontal component −2 ; −2 ; vertical component −1 −1

7 + 11 i 7 + 11 i

−16 − 30 i −16 − 30 i

−4 − i 10 −4 − i 10

x = 7 − 3 i x = 7 − 3 i

x = −1 , −5 x = −1 , −5

x = 0 , 9 7 x = 0 , 9 7

x = 10 , −2 x = 10 , −2

x = − 1 ± 5 4 x = − 1 ± 5 4

x = 2 5 , − 1 3 x = 2 5 , − 1 3

x = 5 ± 2 7 x = 5 ± 2 7

x = 0 , 256 x = 0 , 256

x = 0 , ± 2 x = 0 , ± 2

x = 11 2 , −17 2 x = 11 2 , −17 2

[ − 10 3 , 2 ] [ − 10 3 , 2 ]

( − 4 3 , 1 5 ) ( − 4 3 , 1 5 )

Where the blue is below the orange line; point of intersection is x = 3.5. x = 3.5.

( 3.5 , ∞ ) ( 3.5 , ∞ )

Practice Test

y = 3 2 x + 2 y = 3 2 x + 2

( 0 , −3 ) ( 0 , −3 ) ( 4 , 0 ) ( 4 , 0 )

( − ∞ , 9 ] ( − ∞ , 9 ]

x = −15 x = −15

x ≠ −4 , 2 ; x ≠ −4 , 2 ; x = − 5 2 , 1 x = − 5 2 , 1

x = 3 ± 3 2 x = 3 ± 3 2

( −4 , 1 ) ( −4 , 1 )

y = −5 9 x − 2 9 y = −5 9 x − 2 9

y = 5 2 x − 4 y = 5 2 x − 4

5 13 − 14 13 i 5 13 − 14 13 i

x = 2 , − 4 3 x = 2 , − 4 3

x = 1 2 ± 2 2 x = 1 2 ± 2 2

x = 1 2 , 2 , −2 x = 1 2 , 2 , −2

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Access for free at https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Algebra and Trigonometry
• Publication date: Feb 13, 2015
• Location: Houston, Texas
• Book URL: https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/algebra-and-trigonometry/pages/chapter-2

© Dec 8, 2021 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.