Donna Kirk

Your Turn

5.1

1.

18 plus 11 OR the sum of 18 and 11

2.

27 times 9 OR the product of 27 and 9

3.

84 divided by 7 OR the quotient of 84 and 7

4.

p

minus

q

OR the difference of

p

and

q

5.2

1.

43 + 67

2.

$\left( {45} \right)\left( 3 \right)$

Alternate answer option:

$45 \cdot 3$

The cross symbol is not usually used in algebra.

\left( {45} \right)\left( 3 \right)

3.

$45 \div 3$

Alternate answer options:

$\frac{{45}}{3}$

45 / 3

45 \div 3

4.

89-42

5.3

1.

n - 20

2.

$6 \left( {n + 2} \right)$

Alternate answer options:

$\left( {n + 2} \right) \cdot 6$

$6 \cdot \left( {n + 2} \right)$

The cross symbol is not usually used in algebra.

\left( {n + 2} \right) \bullet 6

OR

6\left( {n + 2} \right)

3.

${n^3}$	First, cube your variable.
${n^3} - 5$	Then, subtract 5.

{n^3} - 5

4.

60h	First, multiply your variable times 60:
60h + 40	Then, add 40:

60h + 40

5.4

1.

5y = 50

Break your sentence into components.

Five times y	is	50
5y	=	50

$5y = 50$

2.

\frac{1}{2}n = 30

Break your sentence into components.

Half of a number n	is	30
$\frac{1}{2}n$	=	30

$\frac{1}{2}n = 30$

Alternate answer option:

$\frac{n}{2} = 30$

3.

3n - 7 = 2

Break your sentence into components.

The difference of three times a number	and 7	is	2
3n	– 7	=	2

3n – 7 = 2

4.

2x + 7 = 21

Break your sentence into components.

Two times x	plus 7	is	21
2x	+ 7	=	21

$2x + 7 = 21$

5.5

1.

Use PEMDAS to verify.

$24 - (17 - 6) = 24 - 11 = 13$

24 - \left( {17 - 6} \right) = 13

2.

Use PEMDAS to verify.

$\left( {3 \cdot 6} \right) + 13 = 18 + 13 = 31$

\left( {3 \bullet 6} \right) + 13 = 31

3.

Use PEMDAS to verify.

$\left( {12 - 6} \right) \div \left( {5 - 3} \right) = 6 \div 2 = 3$

\left( {12 - 6} \right) \div \left( {5 - 3} \right) = 3

4.

Use PEMDAS to verify.

$5 \cdot \left( {{3^2} + 5} \right) = 5 \cdot \left( {9 + 5} \right) = 5 \cdot 14 = 70$

5 \bullet \left( {{3^2} + 5} \right) = 70

5.6

1.

$5x - 6 = 5(3) - 6$	Substitute.
$15 - 6$	Simplify.

9

9

2.

${x^2} - 6x + 3 = {3^2} - 6\left( 3 \right) + 3$	Substitute.
$9 - 18 + 3$	Simplify.

–6

- 6

5.7

1.

$\left( {2{x^2} - 4x + 5} \right) + \left( {3{x^2} + x - 12} \right)$

Like terms are constants or have the same variable and exponent. Group the like terms.

$\left( {2{x^2} + 3{x^2}} \right) + \left( { - 4x + x} \right) + \left( {5 + \left( { - 12} \right)} \right)$

Add the coefficients and keep the same variable.

$5{x^2} - 3x - 7$

5{x^{ 2}}-3x - 7

5.8

1.

$\left( {{x^2} - 4x + 8} \right) - \left( {{x^2} + 5x + 12} \right)$

{x^2} - 4x + 8 - {x^2} - 5x - 12

Distribute the subtraction.

Like terms are constants or have the same variable and exponent. Group the like terms.

$\left( {{x^2} - {x^2}} \right) + \left( { - 4x - 5x} \right) + \left( {8 - 12} \right)$

Add the coefficients and keep the same variable.

$- 9x - 4$

- 9x – 4

5.9

1.

2y + 10

$2\left( {y + 5} \right)$	Use the distributive property.
$2y + 2(5)$
$2y + 10$

2.

- 2a - 2b + 8

$\left( { - 2} \right)\left( {a + b - 4} \right)$	Use the distributive property.
$\left( { - 2} \right)a + \left( { - 2} \right)b + \left( { - 2} \right)\left( { - 4} \right)$
$- 2a - 2b + 8$

3.

112 + 16x

${4^2}\left( {47 - 40 + x} \right)$	Simplify. Subtract inside the parentheses.
${4^2}\left( {7 + x} \right)$	Evaluate the exponent.
$16\left( {7 + x} \right)$	Use the distributive property.
$16\left( 7 \right) + 16\left( x \right)$
$112 + 16x$

4.

6x + 18

$\left( {18 \div 3} \right)\left( {x + 7 - 4} \right)$	Simplify. Divide in the first parentheses.
$\left( 6 \right)\left( {x + 7 - 4} \right)$	Subtract in the second parentheses.
$\left( 6 \right)\left( {x + 3} \right)$	Use the distributive property.
$6x + 6\left( 3 \right)$
$6x + 18$

5.

$2\left( {3a + 5} \right) + \left( { - 3} \right)\left( {a + 2} \right)$	Use the distributive property.
$2\left( {3a} \right) + 2\left( 5 \right) + \left( { - 3} \right)\left( a \right) + \left( { - 3} \right)\left( 2 \right)$
$6a + 10 - 3a - 6$

Like terms are constants or have the same variable and exponent. Group the like terms.

$\left( {6a - 3a} \right) + \left( {10 - 6} \right)$

Add the coefficients and keep the same variable.

$3a + 4$

3a + 4

5.10

1.

$\left( {x - 4} \right)\left( {2x - 3} \right)$

Use the distributive property.

$x\left( {2x - 3} \right) - 4\left( {2x - 3} \right)$

Use the distributive property again.

$x\left( {2x} \right) - x\left( 3 \right) - 4\left( {2x} \right) - 4\left( { - 3} \right)$

Simplify.

$2{x^2} - 3x - 8x + 12$

Like terms are constants or have the same variable and exponent. Add the coefficients and keep the same variable.

$2{x^2} - 11x + 12$

5.11

1.

$4{x^2} + x - 2$

$\left( {16{x^2} + 4x - 8} \right) \div \left( 4 \right)$	Divide each term by 4.
$\left( {16{x^2} \div 4} \right) + \left( {4x \div 4} \right) + \left( { - 8 \div 4} \right)$	Simplify.
$4{x^2} + x - 2$

5.12

1.

$2\left( {x + 1} \right) - 3 = 5$

Step 1: Simplify each side as much as possible.

$2x + 2 - 3 = 5$

$2x - 1 = 5$

Step 2: Collect the variable terms on one side.

$2x - 1 = 5$

There is nothing to do here.

Step 3: Collect the constant terms on the other side.

$2x - 1 + 1 = 5 + 1$

$2x = 6$

Step 4: Make the coefficient of the variable term equal to 1.

$\frac{{2x}}{3} = \frac{6}{3}$

$x = 3$

Step 5: Checking is left to you.

x = 3

5.13

1.

$6\left( {y - 2} \right) - 5y = 4\left( {y + 3} \right) - 4\left( {y - 1} \right)$

Step 1: Simplify each side as much as possible.

$6y - 12 - 5y = 4y + 12 - 4y + 4$

$y - 12 = 16$

Step 2: Collect the variable terms on one side.

$y - 12 = 16$

There is nothing to do here.

Step 3: Collect the constant terms on the other side.

$y - 12 + 12 = 16 + 12$

$y = 28$

Step 4: Make the coefficient of the variable term equal to 1.

$y = 28$

You are already done. This is the solution.

Step 5: Checking is left to you.

y = 28

5.14

1.

Let S = Sam’s books and H = Henry’s books.

Together they have 68 books, so S + H = 68

Sam has 26 books, so substitute 26 for S.

26 + H = 68

Solve for H.

Step 1: Simplify each side as much as possible.

26 + H = 68

Step 2: Collect the variable terms on one side.

26 + H = 68

This is already done.

Step 3: Collect the constant terms on the other side.

26 + H – 26 = 68 – 26

H = 42

Step 4: Make the coefficient of the variable term equal to 1.

H = 42

You are already done. This is the solution.

Step 5: Checking is left to you.

Henry has 42 books.

5.15

1.

The average price is $3.053 per gallon. Aiko fills up his car with 16 gallons.

Total cost = $16{\rm{\ gallons}}\left( {\frac{{3.053{\rm{\ dollars}}}}{{{\rm{1\ gallon}}}}} \right) \approx 48.85{\rm{\ dollars}}$

The total cost is $48.85.

Total cost is $48.85

5.16

1.

The application you write will vary.

Then, solve the algebraic equation: $25x + 75 = 200$

Step 1: Simplify each side as much as possible.

This is already done.

$25x + 75 = 200$

Step 2: Collect the variable terms on one side.

This is already done.

$25x + 75 = 200$

Step 3: Collect the constant terms on the other side.

$25x + 75 - 75 = 200 - 75$

$25x = 125$

Step 4: Make the coefficient of the variable term equal to 1.

$\frac{{25x}}{{25}} = \frac{{125}}{{25}}$

$x = 5$

Write a sentence that uses the 5 regarding your application.

Step 5: Checking is left to you.

Answers will vary. For example: You can rent a paddleboard for $25 per hour with a water shoe purchase of $75. If you spent $200, how many hours did you rent the paddle board for?
You rented the paddle board for 5 hours.

5.17

1.

$2\left( {x + 6} \right) = 3x + 4 - \left( {x + 5} \right)$

Step 1: Simplify each side as much as possible.

$2x + 12 = 3x + 4 - x - 5$

$2x + 12 = 2x - 1$

Step 2: Collect the variable terms on one side.

$2x + 12 - 2x = 2x - 1 - 2x$

$12 = - 1$

This is a false statement, so this equation has no solution.

12 = - 1

, which is false; therefore, this is a false statement, and the equation has no solution.

5.18

1.

$3x - 7 - \left( {x + 5} \right) = 2\left( {x - 6} \right)$

Step 1: Simplify each side as much as possible.

$3x - 7 - x - 5 = 2x - 12$

$2x - 12 = 2x - 12$

Step 2: Collect the variable terms on one side.

$2x - 12 - 2x = 2x - 12 - 2x$

$- 12 = - 12$

This statement is always true. The variable does not matter at all, so there are infinitely many solutions.

- 12 = - 12

, which is true; therefore, this is a true statement, and there are infinitely many solutions.

5.19

1.

Solve $I = Prt$ for t. Treat t as the variable and everything else as “constants.”

You can skip to Step 4: Make the coefficient of the variable term equal to 1.

$\frac{I}{{Pr}} = \frac{{Prt}}{{Pr}}$

$\frac{I}{{Pr}} = t$

Alternate answer option: $t = \frac{I}{{Pr}}$

Step 5: Checking is left to you.

t = \frac{I}{{\text{P}}r}

5.20

1.

$V = \frac{1}{3}\pi {r^2}h$

Treat h as the variable and everything else as “constants.”

You can skip to Step 4: Make the coefficient of the variable term equal to 1.

$3V = \frac{1}{3}\pi {r^2}h\left( 3 \right)$

$3V = \pi {r^2}h$

$\frac{{3V}}{{\pi {r^2}}} = \frac{{\pi {r^2}h}}{{\pi {r^2}}}$

$\frac{{3V}}{{\pi {r^2}}} = h$

Alternate answer option: $h = \frac{{3V}}{{\pi {r^2}}}$

h = \frac{{3V}}{{\pi {r^2}}}

5.21

1.

For the graph, shade to the left of 2.5 to show the solution includes all numbers left of 2.5, then put a parenthesis at 2.5 to show that 2.5 is not included.

Write in interval notation with a parenthesis and negative infinity on the left to show that the solution includes all numbers less than 2.5. After a comma, write 2.5 with a parenthesis to show that 2.5 is not included in the solution.

(–∞, 2.5)

A number line ranges from 0 to negative 3, in increments of 1. A close parenthesis is marked at 2.5. The region to the left of the parenthesis is shaded on the number line. Text reads, (negative infinity, 2.5)

5.22

1.

Graph $x \ge 0$ and $x \le 2.5$ . The area where they overlap is the solution.

Graphically: Place a bracket at 0 facing to the right and a bracket at 2.5 facing to the left. Shade in between the two brackets.

Interval notation: [0,2.5]

0 \leq x \leq 2.5

Graph:

A number line ranges from negative 1 to 3.5, in increments of 0.5. An open square bracket is marked at 0.0 and a close parenthesis is marked at 2.5. The region within the parentheses is shaded on the number line. Text reads, (0, 2.5)

5.23

1.

First solve the inequality.

$- 13m \ge 65$	Divide by –13.
$\frac{{ - 13m}}{{ - 13}} \le \frac{{65}}{{ - 13}}$	Switch the sign to ≤ since you divided by a negative number.
$m \le - 5$

For the graph, shade to the left of $- 5$ to show all numbers left of $- 5$ are included, then put a bracket at $- 5$ to show that the solution includes $- 5$ .

Write in interval notation with a parentheses and negative infinity on the left to show that the solution includes all numbers less than $- 5$ . After a comma, write $- 5$ with a bracket to show that $- 5$ is included in the solution.

(–∞, –5]

A number line ranges from negative 7 to negative 4, in increments of 1. A close square bracket is marked at negative 5. The region to the left of the square bracket is shaded on the number line. Text reads, m is less than or equal to negative 5, (negative infinity, negative 5)

5.24

1.

$8p + 3\left( {p - 12} \right) \ge 7p - 28$

First solve the inequality.

$8p + 3\left( {p - 12} \right) \ge 7p - 28$	Distribute.
$8p + 3p - 36 \ge 7p - 28$	Simplify.
$11p - 36 \ge 7p - 28$ .
$11p - 36 - 7p \ge 7p - 28 - 7p$	Collect variable terms on the left.
$4p - 36 \ge - 28$
$4p - 36 + 36 \ge - 28 + 36$	Collect constants on the right.
$4p \ge 8$
$\frac{{4p}}{4} \ge \frac{8}{4}$	Divide by 4.
$p \ge 2$

For the graph, place a bracket at 2 to show that the solution includes 2. Shade to the right of $2$ to show all numbers right of $2$ are included.

Write in interval notation with a bracket at 2 to show that 2 is included. Then shade to the right to show that all numbers greater than 2 are part of the solution.

p \geq 2

A number line ranges from 0 to 3, in increments of 1. An open square bracket is marked at 2. The region to the right of the square bracket is shaded on the number line. Text reads, (2, infinity).

5.25

1.

Break the words into components.

The bill	is no more than	$28.80 flat fee	plus	$0.20 per message
B	≥	28.80	+	0.20m

Keep the bill to no more than $50.

Keep $28.80 + 0.2m \le 50$ .

Solve the inequality $28.80 + 0.2m \le 50$ .

$28.80 + 0.2m \le {\rm{ }}50 - 28.80$	Collect the constants on one side.
$0.2m \le {\rm{ }}21.2$
$\frac{{0.2m}}{{0.2}} \le {\rm{ }}\frac{{21.2}}{{0.2}}$	Divide both sides by 0.2.
$m \le 106$

Taleisha can send no more than 106 text messages per month.

Taleisha can send/receive 106 or fewer text messages and keep her monthly bill no more than $50.

5.26

1.

Break the words into components.

The number of credit hours you can take	is no more than	$1,500 saved	plus	$500 scholarship	divided by	$113 cost of one credit hour
H	≤	(1,500	+	500)	/	113

You can spend no more than the money you saved plus your scholarship.

Solve $H \le \frac{{1,500 + 500}}{{113}}$

$H \le \frac{{2,000}}{{113}}$

$H \le 17.699....$

Since you can only take a whole number of credit hours, you can take at most 17 credit hours.

You could take up to 17 credit hours and stay under $2,000.

5.27

1.

Malik must tutor at least 23 hours.

Malik plans a 6-day vacation.

$840 in savings.

Earns $45 per hour.

Static Costs: $525, $780

Daily costs: $95 per night for 6 nights = $570

Savings – Costs: $840-\left( {525 + 780 + 570} \right) = 840-1,875 = -1,035$

Malik still needs to earn $1,035.

Break the words into components.

1,035	is no less than	45	times	hours worked
1,035	≤	45	∙	h

Solve the inequality $1,035 \le 45h$

$1,035 \le 45h$	Divide both sides by 45.
$\frac{{1,035}}{{45}} \le \frac{{45h}}{{45}}$
$23 \le h$

Malik needs to tutor at least 23 hours.

5.28

1.

There are three ways to write ratios: a to b, a:b, or a/b, which can be written with fraction notation $\left( {\frac{a}{b}} \right)$ .

If $a = 1$ U.S. dollar, and $b = 1.21$ Canadian dollars, the ratio can be written in one of three ways.

1 to 1.21

1:1.21

$\frac{1}{{1.21}}$

a

= 1 U.S. dollar, and

b

= 1.21 Canadian dollars, the ratio is 1 to 1.21; or 1:1.21; or

\frac{1}{{1.21}}

.

5.29

1.

There are three ways to write ratios: a to b, a:b, or a/b, which can be written with fraction notation $\left( {\frac{a}{b}} \right)$ .

If $a = 170$ pounds, and $b = 64$ pounds, the ratio can be written in one of three ways.

170 to 64

170:64

$\frac{{170}}{{64}}$

With

a

= 170 pounds on Earth, and

b

= 64 pounds on Mars, the ratio is 170 to 64; or 170:64; or

\frac{{170}}{{64}}

.

5.30

1.

Set up the two ratios as proportions.

Let x = the number of dollars you should receive.

Create ratios with U.S. dollars in the numerator and Euros in the denominator. It helps many students to create a table to help set up their ratios.

	Exchange	What I have or want
U.S. Dollars	1	x
Euros	0.82	180

Set up your proportion.

$\frac{1}{{0.82}} = \frac{x}{{180}}$
$1(180) = 0.82x$	Cross multiply.
$\frac{{180}}{{0.82}} = \frac{{0.82x}}{{0.82}}$	Divide. Approximate to two decimal places since this is money.
$219.51 \approx x$

You should receive $219.51.

$219.51

5.31

1.

Set up the two ratios as proportions.

Let x = weight on the moon.

It helps many students to create a table to help set up their ratios.

	Conversion	What I have or want
Weight on Earth	200	3,040
Weight on the moon	33	x

Set up your proportion.

$\frac{{200}}{{33}} = \frac{{3,040}}{x}$
$200x = \left( {33} \right)\left( {3,040} \right)$	Cross multiply.
$200x = 100,320$
$\frac{{200x}}{{200}} = \frac{{100,320}}{{200}}$	Divide.
$x = 501.6$

The weight of the car on Mars is exactly 501.6 pounds.

501.6 pounds

5.32

1.

Set up the two ratios as proportions.

Let x = the number of cookies you can make.

It helps many students to create a table to help set up their ratios.

	Recipe	What I have or want
Cups of flour	2.25	27
Cookies	60	x

Set up your proportion.

$\frac{{2.25}}{{60}} = \frac{{27}}{x}$
$2.25x = \left( {60} \right)\left( {27} \right)$	Cross multiply.200x
$2.25x = 1,620$
$\frac{{2.25x}}{{2.25}} = \frac{{1,620}}{{2.25}}$	Divide.
$x = 720$

You can make 720 cookies.

720 cookies

5.33

1.

Set up a table to make the comparisons.

	Floor tile	book	table	pencil
Length (inches)	24	13	60	7.5
Length (cm)	60.96	33.02	152.4	19.05
Ratio $\frac{{cm}}{{in}}$	$\frac{{60.96}}{{24}}$	$\frac{{33.02}}{{13}}$	$\frac{{152.4}}{{60}}$	$\frac{{19.05}}{{7.5}}$
Constant Divide to find the constant.	2.54	2.54	2.54	2.54

The constant of proportionality is 2.54. It tells you that 2.54cm is approximately 1 inch.

The constant of proportionality (centimeters divided by inches) is 2.54. This tells you that there are 2.54 centimeters in one inch.

5.34

1.

Zac runs at a constant speed of 4 miles per hour.

Set up the two ratios as proportions to answer this question.

Let x = the time spent running.

It helps many students to create a table to help set up their ratios.

	Exchange	What I have or want
miles	4	122
hours	1	x

Set up your proportion.

$\frac{4}{1} = \frac{{122}}{x}$
$4x = \left( 1 \right)\left( {122} \right)$	Cross multiply
$\frac{{4x}}{4} = \frac{{122}}{4}$	Divide.
$x = 30.5$

Zac ran 30.5 hours.

Alternate answer: Zac ran for 30 hours and 30 minutes.

30.5 hours (or 30½ hours, or 30 hours and 30 minutes)

5.35

1.

Joe is paid $2.50 for each bucket.

Joe filled 83 buckets.

Set up the two ratios as proportions to answer this question.

Let x = the amount of money earned.

It helps many students to create a table to help set up their ratios.

	Conversion	What I have or want
buckets	1	83
dollars	$2.50	x

Set up your proportion.

$\frac{1}{{2.50}} = \frac{{83}}{x}$
$x = \left( {2.50} \right)\left( {83} \right)$	Cross multiply.
$x = 207.50$

Joe earned $207.50.

$207.50

5.36

1.

0.62 miles is approximately 1 kilometer.

Montreal is 203 kilometers away.

Let x = the distance to Montreal in miles.

It helps many students to create a table to help set up their ratios.

	Conversion	What I have or want
kilometers	1	203
miles	0.62	x

Set up your proportion.

$\frac{1}{{0.62}} = \frac{{203}}{x}$
$x = \left( {0.62} \right)\left( {203} \right)$	Cross multiply.
$x \approx 125.9{\rm{ miles}}$

The distance to Montreal is approximately 125.9 miles.

125.9 miles

5.37

1.

The scale will vary for each student, depending on the screen they are using to view the map. These calculations will assume the borders of your map are as follows:

The map’s southern border is 4 inches.

The map’s northern border is also 4 inches.

The map’s eastern and western borders are both 3 inches.

The real southern border is 365 miles.

Let x = the scale factor (how many real miles are equal one inch on the map).

It helps many students to create a table to help set up their ratios.

	Conversion	What I have or want
map inches	4	1
miles	365	x

Set up your proportion.

$\frac{4}{{365}} = \frac{1}{x}$
$4x = \left( {365} \right)\left( 1 \right)$	Cross multiply.
$\frac{{4x}}{4} = \frac{{365}}{4}$	Divide.
$x = 91.25{\rm{\ miles}}$

The scale factor is one map inch equals 91.25 real miles.

Now use the scale to find the length of the eastern and western borders.

Let b = the length of the western border.

	Scale	Western Border
map inches	1	3
miles	91.25	b

$\frac{1}{{91.25}} = \frac{3}{b}$
$b = \left( {91.25} \right)\left( 3 \right)$	Cross multiply.
$b = 273.75{\rm{ miles}}$

The western and southern borders are 273.75 miles long. The northern and southern borders are 365 miles long. The scale is one inch = 91.25 miles.

The scale is

1{\text{ inch}} = 91.25{\text{ miles}}

. The other borders would calculate as: eastern and western borders are 273.75 miles, and northern border is 365 miles.

5.38

1.

Let x = the length of the real Jeep.

	Scale	What I know or want to know
toy Jeep	1	11.5 inches
real Jeep	16	x

$\frac{1}{{16}} = \frac{{11.5}}{x}$
$x = \left( {16} \right)\left( {11.5} \right)$	Cross multiply.

$x = 184{\rm{\ inches = 15\ feet\ 4\ inches}}$

The real Jeep is 184 inches long. You can also express your answer as 15 feet 4 inches or as 15’ 4”.

184 inches, or 15 feet, 4 inches.

5.39

1.

$(–4, 2)$ : For the graph, move 4 units left and 2 units up to plot the point. The point is in quadrant II.

$(–1, –2)$ : For the graph, move 1 unit left and 2 units down to plot the point. The point is in quadrant III.

$(3, –5)$ : For the graph, move 3 units right and 5 units down to plot the point. The point is in quadrant IV.

$(–3, 0)$ : For the graph, move 3 units left and plot the point on the negative x-axis.

$\left( {\frac{5}{3},2} \right)$ : For the graph, move $\frac{5}{3}$ units right and 2 units up to plot the point. The point is in quadrant I.

Five points are marked on an x y coordinate plane. The x and y axes range from negative 8 to 8, in increments of 2. The first, second, third, and fourth quadrants are labeled. The points are plotted at the following coordinates: a (negative 4, 2), b (negative 1, negative 2), c (3, negative 5), d (negative 3, 0), and e (negative 1.5, 2). Note: all values are approximate.

5.40

1.

$y = x + 2$

$(0, 2)$

Is the ordered pair a solution to the equation?

Substitute the pair into the equation.

$2 = 0 + 2$

This is true, so the ordered pair is a solution.

Is the point on the line?

Yes. If the ordered pair is a solution to the equation, then the ordered pair is a point on the line.

Yes
Yes

2.

$(1, 2)$

Is the ordered pair a solution to the equation? Substitute the pair into the equation.

$2 = 1 + 2$

This is not true, so the ordered pair is not a solution.

Is the point on the line?

No. If the ordered pair is not a solution to the equation, then the ordered pair is not a point on the line.

No
No

3.

$(–1, 1)$

Is the ordered pair a solution to the equation? Substitute the pair into the equation.

$1 = –1 + 2$

This is true, so the ordered pair is a solution.

Is the point on the line?

Yes. If the ordered pair is a solution to the equation, then the ordered pair is a point on the line.

Yes
Yes

4.

$(–3, –1)$

Is the ordered pair a solution to the equation? Substitute the pair into the equation.

$–1 = –3 + 2$

This is true, so the ordered pair is a solution.

Is the point on the line?

Yes. If the ordered pair is a solution to the equation, then the ordered pair is a point on the line.

Yes
Yes

5.41

1.

$y = 3x - 1$

Find three points that are solutions to the equation. Choose three values for x.

x	y
–1
0
1

Substitute into the equation to find the values for y.

x	y	$y=3x–1$
–1	–4	$3(–1) – 1 = –3 – 1 = –4$
0	–1	$3(0) – 1 = 0 – 1 = –1$
1	2	$3(1) – 1 = 3 – 1 = 2$

Plot the points. Check that they line up and draw the line.

A line is plotted on an x y coordinate plane. The x and y axes range from negative 12 to 12, in increments of 2. The line passes through the following points, (negative 2, negative 7), (0, negative 1), (2, 5), and (4, 11). Note: all values are approximate.

5.42

1.

Break the words into components.

Let x = the number stamps.

Let y = the cost of mailing your bundle of letters.

Cost of mailing	is	cost of one stamp	times	number of letters
y	=	$0.55	∙	x

$y = 0.55x$
$y = 0.55(3)$	Substitute 3 for x.
$y = 1.65$

Your friend will pay $1.65 for stamps.

A line is plotted on a coordinate plane. The horizontal axis ranges from 0 to 100, in increments of 10. The vertical axis ranges from 0 to 50, in increments of 10. A stamp of the USA flag is at the top-left. The line passes through the points, (3, 1.65), (20, 11), and (100, 55).

Your friend will pay $1.65.

5.43

1.

$y > x - 1$

$(0, 1)$

Is the ordered pair a solution to the inequality?

Substitute the pair into the inequality.

$1 > 0 - 1$

$1 > -1$

This is true, so the ordered pair is a solution.

Yes

2.

$y > x - 1$

$(-4, -1)$

Is the ordered pair a solution to the inequality?

Substitute the pair into the inequality.

$-1 > -4 - 1$

$-1 > -5$

This is true, so the ordered pair is a solution.

Yes

3.

$y > x - 1$

$(4, 2)$

Is the ordered pair a solution to the inequality?

Substitute the pair into the inequality.

$2 > 4 - 1$

$2 > 3$

This is not true, so the ordered pair is not a solution.

No

4.

$y > x - 1$

$(3, 0)$

Is the ordered pair a solution to the inequality?

Substitute the pair into the inequality.

$0 > 3 - 1$

$0 > 2$

This is not true, so the ordered pair is not a solution.

No

5.

$y > x - 1$

$(-2, -3)$

Is the ordered pair a solution to the inequality?

Substitute the pair into the inequality.

$-3 > -2 - 1$

$-3 > -3$

This is not true, so the ordered pair is not a solution.

No

5.44

1.

The boundary line for this graph is given: $y = - 2x + 3$ .

Test a point not on the line to see if $y \ge - 2x + 3$ is the answer:

$(0, 0)$

$0 \ge - 2\left( 0 \right) + 3$

$0 \ge 3$ is not true, so $(0, 0)$ is not a solution. That is good, since the proposed solution is the points above the given line.

Test $(5,1)$ .

$1 \ge - 2\left( 5 \right) + 3$

$1 \ge -10 + 3$

$1 \ge -7$ is true, so shading above the line is correct.

$y \ge - 2x + 3$ is the answer.

y \geq - 2x + 3

5.45

1.

A dashed line is plotted on an x y coordinate plane. The x and y axes range from negative 10 to 10, in increments of 2.5. The line passes through the following points, (negative 10, negative 7.5), (0, negative 1), (5, 2.5), and (7.5, 3.75). The region above the line is shaded. Note: all values are approximate.

Graph the line $y = \frac{2}{3}x - 1$ with a dotted line since the inequality is strictly greater than.

Shade above the line since this is “y >”.

Test $(0,0)$ to be sure shading above the line is correct.

$0 > \frac{2}{3}\left( 0 \right) - 1$

$0 > -1$ is true, so $(0,0)$ is in the solution. The shading should include $(0, 0)$ . The shading is correct.

5.46

1.

11x + 16.5y \geq 330

Break the words into components.

Let x = the number of hours worked at the gas station.

$11 = the amount paid per hour at the gas station.

Let y = the number of hours worked as an IT consultant.

$16.50 = the amount paid per hour as an IT consultant.

$330 = the least amount needed to earn per week.

Gas station pay	plus	IT pay	is at least	330
11x	+	16.50y	≥	330

Write an inequality to model the situation:

$11x + 16.50y \ge 330$

2.

A line is plotted on an x y coordinate plane. The x-axis ranges from 0 to 35, in increments of 5. The y-axis ranges from 0 to 35, in increments of 5. The line passes through the following points, (0, 20), (15, 10), and (30, 0). The region above the line is shaded.

Graph the inequality: $11x + 16.50y \ge 330$ .

Graph the line $11x + 16.50y = 330$ with a solid line since it is an “or equal to” type.

Test $(0, 0)$ in $11x + 16.50y \ge 330$ .

$11\left( 0 \right) + 16.50\left( 0 \right) \ge 330$

$0 \ge 330$ is false, so shade the other side of the boundary.

The shading should not include $(0, 0)$ .

3.

Answers will vary.

Test $(0, 0)$ in $11x + 16.50y \ge 330$ .

$11\left( 0 \right) + 16.50\left( 0 \right) \ge 330$

$0 \ge 330$ is false. It is not part of the solution.

Not working at all would not be good for Harrison. He would earn no money.

Test $(15, 5)$ .

$11\left( {15} \right) + 16.50\left( 5 \right) \ge 330$

$247.5 \ge 330$

This is not part of the solution.

If Harrison works at the gas station for 15 hours and at the IT job for 5 hours, that will not earn enough money.

Test $(30,10)$ .

$11\left( {30} \right) + 16.50\left( {10} \right) \ge 330$

$495 \ge 330$

This point is part of the solution.

If Harrison works at the gas station for 30 hours and 10 hours at the IT job, that will earn $496. That is more than enough money to exceed the $330 requirement. This is still working just 40 hours a week and is very doable.

5.47

1.

You can use a table to help you practice distribution until you get comfortable with multiplying polynomials. Distribution works no matter how many terms you have. You can always make a table like this if you have problems keeping track of the terms.

	x	+ 3
x	${x^2}$	3x
+ 1	x	3

After multiplying, write the products from the cells as a sum.

${x^2} + 3x + x + 3$

Combine like terms.

${x^2} + 4x + 3$

{x^2} + 4x + 3

5.48

1.

2{x^2} - 5x - 3

You can use a table to help you practice distribution until you get comfortable with multiplying polynomials. Distribution works no matter how many terms you have. You can always make a table like this if you have problems keeping track of the terms.

	x	– 3
2x	$2{x^2}$	–6x
+ 1	x	–3

After multiplying, write the products from the cells as a sum.

$2{x^2} - 6x + x - 3$

Combine like terms.

$2{x^2} - 5x - 3$

5.49

1.

\left( x + 2 \right)\left( x + 4 \right)

${x^2} + 6x + 8$

Look for factors of 8 that add up to 6. This works when the coefficient of the ${x^2}$ term is 1.

Factors of 8	Sum of Factors = 6	Success?
1 times 8	$1 + 8 = 9$	No
2 times 4	$2 + 4 = 6$	Yes!

Because 2 and 4 are factors of 8 and they add up to 6, you can make the factors of the polynomial.

$(x + 2)(x + 4)$

5.50

1.

${x^2} - 16x + 63$

Look for factors of 63 that add up to $-16$ . This works when the coefficient of the ${x^2}$ term is 1.

Factors of 63	Sum of Factors = –16	Success?
1 times 63	$1 + 63 = 64$	No
3 times 21	$3 + 21 = 24$	No
7 times 9	$7 + 9 = 16$	No, but it’s the opposite of the sum you want. Try negatives of your factors.
–7 times –9	$-7 + \left({-9} \right) = -16$	Yes

Because –7 and –9 are factors of 63 and they add up to –16, you can make the factors of the polynomial.

$(x - 7)(x - 9)$

\left( {x - 7} \right)\left( {x - 9} \right)

5.51

1.

$y = - {x^2}$

For now, solve by plotting several points.

$x$	$y$	$y = - {x^2}$
–3	–9	$y = - {\left({ - 3} \right)^2}$
–2	–4	$y = - {\left({ - 2} \right)^2}$
–1	–1	$y = - {\left({ - 1} \right)^2}$
0	0	$y = - {\left(0 \right)^2}$
1	–1	$y = - {\left(1 \right)^2}$
2	–4	$y = - {\left(2 \right)^2}$
3	–9	$y = - {\left(3 \right)^2}$

Connect the points with a smooth curve.

A parabola is plotted on an x y coordinate plane. The x and y axes range from negative 10 to 10, in increments of 1. The parabola opens down and it passes through the following points, (negative 2, negative 4), (negative 1, negative 1), (0, 0), (1, negative 1), and (2, negative 4). Note: all values are approximate.

5.52

1.

y = 0

where

x = -3

or

x = 2

x = - 3,\,\,x = 2

5.53

1.

$\left({x - 2} \right)\left({x + 5} \right) = 0$

$x - 2 = 0$	$x + 5 = 0$	Set factors equal to zero.
$x = 2$	$x = - 5$	Solve each new equation.

Checking is left to you.

x = 2

or

x = - 5

5.54

1.

${x^2} + 2x - 15 = 0$
$\left({x - 3} \right)\left({x + 5} \right) = 0$		Factor the polynomial.
$x - 3 = 0$	$x + 5 = 0$	Set factors equal to zero.
$x = 3$	$x = - 5$	Solve each new equation.

Checking is left to you.

x = 3

or

x = - 5

5.55

1.

${x^2} = 25$
$x = \pm \sqrt {25}$		Use the Square Root Property.
$x = \pm 5$		Simplify.
$x = 5$	$x = - 5$	Rewrite the two solutions.

Checking is left to you.

x = 5

,

x = -5

5.56

1.

$25{p^2} = 49$
${p^2} = \frac{{49}}{{25}}$		Divide.
$p = \pm \sqrt {\frac{{49}}{{25}}}$		Use the Square Root Property.
$p = \pm \frac{7}{5}$		Simplify.
$p = \frac{7}{5}$	$p = - \frac{7}{5}$	Rewrite the two solutions.

Checking is left to you.

p = \frac{7}{5}

,

p = - \frac{7}{5}

5.57

1.

{a^2} - 2a - 15 = 0

The equation is in standard form.

For the quadratic formula, $a = 1$ , $b = -2$ , and $c = -15$ . Do not get confused because there are two a’s in this exercise.

The solutions to the equation can be found with the quadratic formula $\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .

$\frac{{ - \left({ - 2} \right) \pm \sqrt {{{\left({ - 2} \right)}^2} - 4\left(1 \right)\left({ - 15} \right)} }}{{2\left(1 \right)}}$		Substitute.
$\frac{{2 \pm \sqrt {4 + 60} }}{2}$		Simplify.
$\frac{{2 \pm \sqrt {64} }}{2}$
$\frac{{2 \pm 8}}{2}$
$a = \frac{{2 + 8}}{2}$	$a = \frac{{2 - 8}}{2}$	Rewrite to show two solutions.
$a = \frac{{10}}{2}$	$a = \frac{{ - 6}}{2}$	Simplify.
$a = 5$	$a = - 3$

Checking is left for you.

a = 5

,

a = - 3

5.58

1.

3{y^2} - 5y + 2 = 0

The equation is in standard form.

For the quadratic formula, $a = 3$ , $b = -5$ , and $c = 2$ .

The solutions to the equation can be found with the quadratic formula $\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .

$\frac{{ - \left({ - 5} \right) \pm \sqrt {{{\left({ - 5} \right)}^2} - 4\left(3 \right)\left(2 \right)} }}{{2\left(3 \right)}}$		Substitute.
$\frac{{5 \pm \sqrt {25 - 24} }}{6}$		Simplify.
$\frac{{5 \pm \sqrt 1 }}{6}$
$\frac{{5 \pm 1}}{6}$
$y = \frac{{5 + 1}}{6}$	$y = \frac{{5 - 1}}{6}$	Rewrite to show two solutions.
$y = \frac{6}{6}$	$y = \frac{4}{6}$	Simplify.
$y = 1$	$y = \frac{2}{3}$

Checking is left for you.

y = 1

,

y = \frac{2}{3}

5.59

1.

Let $y = {\rm{one}}\;{\rm{integer}}.$

Let $y + 1 = {\rm{the}}\;{\rm{next}}\;{\rm{integer}}.$

Break the sentence into components.

The product of two consecutive integers			is	240
$y$	multiplied by	$y + 1$	=	240

$y\left({y + 1} \right) = 240$	The equation is not in standard form.
${y^2} + y = 240$	Distribute on the left.
${y^2} + y - 240 = 240 - 240$	Subtract to bring it into standard form.
${y^2} + y - 240 = 0$	The equation is in standard form. Use one of the solution methods.

Hint: Use https://www.wolframalpha.com/ to “factor 240” and look for a pair whose sum is 1 and whose product is 240. There is one pair that works: 16 and –15. You want the signs different to have a negative product. You want the 16 positive to have the middle term positive.

$\left({y + 16} \right)\left({y - 15} \right) = 0$
$y + 16 = 0$	$y - 15 = 0$	Set factors equal to zero.
$y = - 16$	$y = 15$	Solve each new equation.
$y + 1 = - 15$	$y + 1 = 16$	Find the second integer for each solution.

There are two pairs of numbers that work: –16 and –15 are one pair; 15 and 16 are the other pair

Checking is left to you.

16, 15 and –16, –15

5.60

1.

Let $w = {\rm{the}}\;{\rm{width}}\;{\rm{of}}\;{\rm{the}}\;{\rm{sign}}\;{\rm{in}}\;{\rm{feet}}.$

Let $w + 1 = {\rm{the}}\;{\rm{length}}\;{\rm{of}}\;{\rm{the}}\;{\rm{sign}}.$

${\rm{Area}} = {\rm{length}}\;{\rm{times}}\;{\rm{width}} = 30\;{\rm{square}}\;{\rm{feet}}.$

Break the sentence into components.

Length	times	width	is	area
$w$	times	$w + 1$	=	30

$w\left({w + 1} \right) = 30$		The equation is not in standard form.
${w^2} + w = 30$		Distribute on the left.
${w^2} + w - 30 = 30 - 30$		Subtract to bring it into standard form.
${w^2} + w - 30 = 0$		The equation is in standard form. Use one of the solution methods.
$\left({w + 6} \right)\left({w - 5} \right) = 0$		This solves by factoring.
$w + 6 = 0$	$w - 5 = 0$	Set factors equal to zero.
$y = - 6$	$w = 5$	Solve each new equation.
$w + 1 = - 5$	$w + 1 = 6$	Find the second integer for each solution.

There is only one pair of numbers that work since the length and width cannot be negative: the width is 5 feet, and the length is 6 feet.

This checks because it yields an area of 30 feet.

{\text{width}} = 5{\text{ feet}}

,

{\text{length}} = 6{\text{ feet}}

5.61

1.

$f\left( x \right) = 3{x^2} - 2x + 1$
$f\left( 3 \right) = 3{\left( 3 \right)^2} - 2\left( 3 \right) + 1$	Substitute 3 for x.
$f\left( 3 \right) = 27 - 6 + 1$	Simplify.
$f\left( 3 \right) = 22$

22

2.

$f\left( x \right) = 3{x^2} - 2x + 1$
$f\left( { - 1} \right) = 3{\left( { - 1} \right)^2} - 2\left( { - 1} \right) + 1$	Substitute –1 for x.
$f\left( 1 \right) = 3 + 2 + 1$	Simplify.
$f\left( 1 \right) = 6$

6

3.

$f\left( x \right) = 3{x^2} - 2x + 1$
$f\left( t \right) = 3{\left( t \right)^2} - 2\left( t \right) + 1$	Substitute t for x.
$f\left( t \right) = 3{t^2} - 2t + 1$	Simplify.

3{t^2} - 2t + 1

5.62

1.

$N\left( t \right) = 100 + 15t$
$N\left( 5 \right) = 100 + 15\left( 7 \right)$	Substitute 7 for t.
$N\left( 5 \right) = 100 + 105$	Simplify.
$N\left( 5 \right) = 205$

Bryan had 205 unread emails after seven days.

There 205 unread emails after 7 days.

5.63

1.

This relation is a function because there are no repeated x-values. Each x-value is matched with only one y-value.

This relation is a function.

2.

This relation is not a function because there are repeated x-values. For example, the x-value 2 is matched with more than one y-value.

This relation is not a function.

5.64

1.

This relation is not a function because there are repeated x-values. Both George and Mike are matched with more than one y-value.

Both George and Mike have two phone numbers. Each

x

-value is not matched with only one

y

-value. This relation is not a function.

5.65

1.

This is the equation of a line. Each x-value on a line is paired with only one y-value.

function

2.

$x + {y^2} = 1$	Isolate the y term.
$x + {y^2} - x = 1 - x$
${y^2} = 1 - x$
$y = \pm \sqrt {1 - x}$	Use the Square Root Property.

Substitute 0 for x.

y = 1 and –1.

Thus, you have two y-values paired with one x-value. This is not a function.

not a function

3.

$y - {x^2} = 2$

$y = {x^2} + 2$

You square x, then add 2. Since each value of x will correspond to just one value of y, this is a function.

function

5.66

1.

This is a function. A vertical line will only pass through it once, so it passes the vertical line test.

This graph represents a function.

5.67

1.

This is not a function. There are places where a vertical line can pass through it more than once. It fails the vertical line test.

This graph does not represent a function.

5.68

1.

The domain is the set of all

x

-values of the relation:

\left\{ {1,2,3,4,5} \right\}

.

2.

The range is the set of all

y

-values of the relation:

\left\{ {1,8,27,64,125} \right\}

.

5.69

1.

The ordered pairs of the relation are:

\left\{ {\left( {-3,\,3} \right),\left( {-2,\,2} \right),\left( {-1,\,0} \right),\left( {0,\,-1} \right),\left( {2,\,-2} \right),\left( {4,\,-4} \right)} \right\}

.

2.

The domain is the set of all

x

-values of the relation:

\left\{ {-3,-2,-1,0,2,4} \right\}

.

3.

The range is the set of all

y

-values of the relation:

\left\{{-4,-2,-1,0,2,3}\right\}

.

5.70

1.

The graph crosses the

x

-axis at the point (2, 0). The

x

-intercept is (2, 0). The graph crosses the

y

-axis at the point (0, −2). The

y

-intercept is (0, −2).

The y-coordinate of the point where the line crosses the x-axis is zero. For this line, that point is (2, 0). The point where the line crosses the x-axis has the form $\left( {a,0} \right)$ and is called the x-intercept of the line. For this line, the x-intercept is (2,0).

In each line, the x-coordinate of the point where the line crosses the y-axis is zero. For this line, that point is (0, –2). The point where the line crosses the y-axis has the form $\left( {0,b} \right)$ and is called the y-intercept of the line. The y-intercept occurs where x is zero. For this line, the y-intercept is (0, –2).

5.71

1.

$3x + y = 12$

Let y = 0 to find the x-intercept.

$3x + 0 = 12$

$3x = 12$

$x = 4$

x-intercept: (4, 0).

Let x = 0 to find the y-intercept.

$3\left( 0 \right) + y = 12$

$y = 12$

y-intercept: (0, 12).

Plot the intercepts and connect to graph the line.

The

x

-intercept is (4, 0) and the

y

-intercept is (0, 12).
A line is plotted on an x y coordinate plane. The x and y axes range from negative 10 to 10, in increments of 1. The line passes through the following points, (0, 12), (4, 0), and (5, negative 3).

A line is plotted on an x y coordinate plane. The x and y axes range from negative 10 to 10, in increments of 1. The line passes through the following points, (0, 12), (4, 0), and (5, negative 3).

5.72

1.

Locate two points on the graph whose coordinates are integers. This solution uses (0, –2) and (3, –6).

Sketch a right triangle on the graph to help you count the rise over run.

Start at (0, –2).

Count the rise. Since the rise goes down, it is negative. The rise is –4.

Count the run. The run goes to the right, so it is positive. The run is 3.

Use the slope formula. $m = \frac{{rise}}{{run}} = \frac{{ - 4}}{3} = - \frac{4}{3}$

The slope of the line is $- \frac{4}{3}$ .

m = \frac{{ - 4}}{3}

5.73

1.

Point 1: (–3, 4) and Point 2: (2, –1).

Find the slope.

$m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{2 - \left( { - 3} \right)}}{{ - 1 - 4}}$	Substitute.
$m = \frac{5}{{ - 5}} = - 1$	Simplify.

The slope is –1.

You can plot the points and verify the slope by counting the rise over the run.

{\text{slope}} = - 1

5.74

1.

When the equation is in the slope-intercept form of the equation of the line,

$y = mx + b$ .

m = the slope.

(0, b) = the y-intercept.

y = 2x - 1

Since this equation is in slope intercept form, the slope is 2.

The y-intercept is (0, –1).

slope

m = 2

and

y{\text{-intercept}} = (0,-1)

2.

When the equation is in the slope-intercept form of the equation of the line,

$y = mx + b$ .

m = the slope.

(0, b) = the y-intercept.

$x + 4y = 16$	Solve for y.
$x + 4y - x = 16 - x$	Subtract.
$4y = - x + 16$
$\frac{{4y}}{4} = \frac{{ - x + 16}}{4}$	Divide.
$y = - \frac{1}{4}x + 4$	This is now in slope intercept form.

The slope is $- \frac{1}{4}$ .

The y-intercept is (0, 4).

slope

m = - \frac{1}{4}

and

y{\text{-intercept}} = (0,4)

5.75

1.

When the equation is in the slope-intercept form of the equation of the line,

$y = mx + b$ .

m = the slope.

(0, b) = the y-intercept.

y = - x - 3

This is in slope intercept form.

The slope is –1. To give you a rise over run, write this as $\frac{{ - 1}}{1}$ .

The y-intercept is (0, –3).

Plot the point (0, –3).

From that point, use the slope $\left( {m = \frac{{rise}}{{run}} = \frac{{ - 1}}{1}} \right)$ to go down one, then right one to plot the next point.

Connect the two points with a line.

5.76

1.

A single variable is a special case.

Make a table of values.

x	y
5	2
5	4
5	–1

Plot the points and connect them with a line.

A line is plotted on an x y coordinate plane. The x and y axes range from negative 10 to 10, in increments of 1. The line is vertical and it passes through the following points, (5, negative 2), (5, 0), and (5, 2). Note: all values are approximate.

5.77

1.

A single variable is a special case.

Make a table of values.

x	y
–3	–4
0	–4
2	–4

Plot the points and connect them with a line.

A line is plotted on an x y coordinate plane. The x and y axes range from negative 10 to 10, in increments of 1. The line is horizontal and it passes through the following points, (negative 2, negative 4), (0, negative 4), and (2, negative 4). Note: all values are approximate.

5.78

1.

(0, 20) is the

y\text{-intercept}

and represents that there were 20 teachers at Jones High School in 1990. There is no

x\text{-intercept}

.

2.

Between 1990 and 1995:

The rise is 10. The run is 5.

Use the slope formula. $m = \frac{{rise}}{{run}} = \frac{{10}}{5} = 2$

On average, the school gained two teachers every year between 1990 and 1995.

Between 1995 and 2000:

The rise is 50 – 30 = 20.

The run is 2000 – 1995 = 5.

Use the slope formula. $m = \frac{{rise}}{{run}} = \frac{{20}}{5} = 4$

On average, the school gained four teachers every year between 1995 and 2000.

Between 2000 and 2005:

The rise is 0.

The run is 5.

The slope is 0.

On average, the school gained zero teachers every year between 2000 and 2005. That means the number of teachers remained constant.

Between 2005 and 2010:

The rise is –10.

The run is 5.

Use the slope formula. $m = \frac{{rise}}{{run}} = \frac{{ - 10}}{5} = - 2$

On average, the school lost two teachers every year between 2005 and 2010.

Between 2010 and 2015:

The rise is 20.

The run is 5.

Use the slope formula. $m = \frac{{rise}}{{run}} = \frac{{20}}{5} = 4$

On average, the school gained four teachers every year between 2010 and 2015.

Between 2015 and 2020:

The rise is 20.

The run is 5.

Use the slope formula. $m = \frac{{rise}}{{run}} = \frac{{20}}{5} = 4$

On average, the school gained four teachers every year between 2015 and 2020, the same slope as between 2010 and 2015.

In the first 5 years the slope is 2; this means that on average, the school gained 2 teachers every year between 1990 and 1995. Between 1995 and 2000, the slope is 4; on average the school gained 4 teachers every year. Then the slope is 0 between 2000 and 2005 meaning the number of teachers remained the same. There was a decrease in teachers between 2005 and 2010, represented by a slope of –2. Finally, the slope is 4 between 2010 and 2020, which indicates that on average the school gained 4 teachers every year.

3.

Answers will vary. Jones High School was founded in 1990 and hired 2 teachers per year until 1995, when they had an increase in students and they hired 4 teachers per year for the next 5 years. Then there was a hiring freeze, and no teachers were hired between 2000 and 2005. After the hiring freeze, the student population decreased, and they lost 2 teachers per year until 2010. Another surge in student population meant Jones High School hired 4 new teachers per year until 2020 when they had 80 teachers at the school.

5.79

1.

$h = 2s + 50$

$h = 2\left( 0 \right) + 50 = 0 + 50 = 50$

The estimated height is 50 inches.

50 inches

2.

$h = 2s + 50$

$h = 2\left( 8 \right) + 50 = 16 + 50 = 66$

The estimated height is 66 inches.

66 inches

3.

For every increase in shoe size, the estimated height increases by two inches. The intercept says that for a shoe size of 0, the estimated height is 50 inches.

The slope, 2, means that the height

h

increases 2 inches when the shoe size(s) increases 1 size.

4.

Plot the point (0, 50). Use the slope

\left( {m = 2 = \frac{{rise}}{{run}} = \frac{2}{1}} \right)

to find another point. From (0, 50), count up 2 and right 1 to determine another point. Plot this point. Connect the two points with a line.

The

h

-intercept means that when the shoe size is 0, the woman’s height is 50 inches. A line is plotted on an x y coordinate plane. The x and y axes range from negative 20 to 140, in increments of 20. The line passes through the points, (negative 30, 0), (0, 50), and (40, 130). Note: all values are approximate.

A line is plotted on an x y coordinate plane. The x and y axes range from negative 20 to 140, in increments of 20. The line passes through the points, (negative 30, 0), (0, 50), and (40, 130). Note: all values are approximate.

5.80

1.

$C = 4p + 25$

$C = 4\left( 0 \right) + 25 = 25$

Stella’s costs are $25 when no pizzas are sold.

$25

2.

$C = 4\left( {15} \right) + 25 = 60 + 25 = 85$

Stella’s costs are $85.

$85

3.

When the equation is in the slope-intercept form of the equation of the line,

$y = mx + b$ .

m = the slope.

(0, b) = the y-intercept.

$C = 4p + 25$

The y-intercept gives the cost when no pizzas are sold: $25.

The slope gives the rate at which costs increase per pizza made. For every pizza sold, Stella’s costs increase by $4.

The slope, 4, means that the weekly cost,

C

, increases by $4 when the number of pizzas sold,

p

, increases by 1. The

C

-intercept means that when the number of pizzas sold is 0, the weekly cost is $25.

4.

Plot the point (0, 25).

From that point, use the slope $\left( {m = \frac{{rise}}{{run}} = \frac{4}{1}} \right)$ to go up 4, then right 1 to plot the next point.

Connect the two points with a line.

Graph:

A line is plotted on an x y coordinate plane. The x-axis ranges from negative 60 to 210, in increments of 15. The y-axis ranges from negative 75 to 200, in increments of 15. The line passes through the points, (negative 15, negative 35), (0, 25), (13, 85), and (30, 145). Note: all values are approximate.

5.81

1.

$\left\{ {\begin{array}{*{20}{c}} {x - 3y = - 8}\\ { - 3x - y = 4} \end{array}} \right.$
Test (2, –2).
$2 - 3\left( { - 2} \right) = - 8$	Substitute into the first equation.
$2 + 6 = - 8$	This is not true. The ordered pair is not a solution to the system because the ordered pair is not a solution to the equation.

No

2.

$\left\{ {\begin{array}{*{20}{c}} {x - 3y = - 8}\\ { - 3x - y = 4} \end{array}} \right.$
Test (–2, 2).
$- 2 - 3\left( 2 \right) = - 8$	Substitute into the first equation.
$- 2 - 6 = - 8$	This is true.
$- 3\left( { - 2} \right) - \left( 2 \right) = 4$	Substitute into the second equation.
$6 - 2 = 4$
$4{\text{ }} = {\text{ }}4$	This is true. This ordered pair is a solution to the system.

Yes

5.82

1.

$\left\{ {\begin{array}{*{20}{c}} {y = \frac{2}{3}x + 1}\\ { - 2x + 3y = 5} \end{array}} \right.$
Test (3, 3).
$3 = \frac{2}{3}\left( 3 \right) + 1$	Substitute into the first equation.
$3 = 2 + 1$	This is true.
$- 2\left( 3 \right) + 3\left( 3 \right) = 5$	Substitute into the second equation.
$- 6 + 9 = 5$
${\text{3 = 5}}$	This is not true. This ordered pair is not a solution to the system.

No

2.

$\left\{ {\begin{array}{*{20}{c}} {y = \frac{2}{3}x + 1}\\ { - 2x + 3y = 5} \end{array}} \right.$
Test (–3, 3)
$3 = \frac{2}{3}\left( { - 3} \right) + 1$	Substitute into the first equation.
$3 = - 2 + 1$	This is not true. This ordered pair is not a solution to the system. There is no need to test the second equation.

No

5.83

1.

Graph the first line by using intercepts: $\left( { - 3,0} \right)$ and $\left( {0,1} \right)$ .

Graph the second line by using intercepts $\left( {5,0} \right)$ and $\left( {0,5} \right)$ .

The two lines cross at $\left( {3,2} \right)$ .

Checking is left to you.

Two lines are plotted on an x y coordinate plane. The x and y axes range from negative 10 to 10, in increments of 1. The first line passes through the points, (negative 9, negative 2), (negative 3, 0), (0, 1), and (9, 4). The second line passes through the points, (negative 4, 9), (0, 5), (5, 0), and (9, negative 4). The two lines intersect at (3, 2).

5.84

1.

$\left\{ {\begin{array}{*{20}{c}} {x - 3y = - 3}\\ {x + y = 5} \end{array}} \right.$

Step 1: Solve one of the equations for either variable.

$x - 3y = - 3$

$x - 3y + 3y = - 3 + 3y$

$x = - 3 + 3y$

Step 2: Substitute the expression from Step 1 into the other equation.

$x + y = 5$

$- 3 + 3y + y = 5$

Step 3: Solve the resulting equation.

$- 3 + 4y = 5$

$- 3 + 4y + 3 = 5 + 3$

$4y = 8$

$y = 2$

Step 4: Substitute the solution in Step 3 into either of the original equations to find the other variable.

$x + y = 5$

$x + 2 = 5$

$x = 3$

Step 5: Write the solution as an ordered pair.

$\left( {3,2} \right)$

Step 6: Checking is left to you.

(3, 2)

5.85

1.

$\left\{ {\begin{array}{*{20}{c}} {x - 3y = - 3}\\ {x + y = 5} \end{array}} \right.$

Step 1: Write both equations in standard form, $Ax + By = C$ .

This system already is in standard form.

Step 2: Make the coefficients of one variable opposites.

$\left( { - 1} \right)\left( {x + y} \right) = \left( { - 1} \right)5$

$- x - y = - 5$

Step 3: Add the equations resulting from Step 2 to eliminate one variable.

	x	–3y	=	–3
	–x	–y	=	–5
SUM:		–4y	=	–8

Step 4: Solve for the remaining variable.

$- 4y = - 8$

$y = 2$

Step 5: Substitute the solution from Step 4 into one of the original equations. Solve for the remaining variable.

$x + y = 5$

$x + 2 = 5$

$x = 3$

Step 6: Write the solution as an ordered pair.

$\left( {3,2} \right)$

Step 6: Checking is left to you.

(3, 2)

5.86

1.

$\left\{ {\begin{array}{*{20}{c}} {y = \frac{1}{3}x + 2}\\ {x - 3y = 7} \end{array}} \right.$

Step 1: Solve one of the equations for either variable.

$y = \frac{1}{3}x + 2$

Step 2: Substitute the expression from Step 1 into the other equation.

$x - 3y = 7$

$x - 3\left( {\frac{1}{3}x + 2} \right) = 7$

$x - x - 6 = 7$

Step 3: Solve the resulting equation.

$- 6 = 7$

This is false. There is no solution to this system.

There is no solution to this system.

5.87

1.

$\left\{ {\begin{array}{*{20}{c}} {y = 3x - 5}\\ { - 18x + 6y = - 30} \end{array}} \right.$

Step 1: Solve one of the equations for either variable.

$y = 3x - 5$

Step 2: Substitute the expression from Step 1 into the other equation.

$- 18x + 6y = - 30$

$- 18x + 6\left( {3x - 5} \right) = - 30$

Step 3: Solve the resulting equation.

$- 18x + 6\left( {3x - 5} \right) = - 30$

$- 18x + 18x - 30 = - 30$

$- 30 = - 30$

This is always true. This system has infinitely many solutions.

There are infinitely many solutions to this system.

5.88

1.

Let e = the number of calories for each minute of elliptical training.

Let c = the number of calories for each minute of circuit training.

Make a table of what you know.

	Elliptical Minutes	Circuit Minutes	Calories
First pass	10	20	278
Second pass	20	30	473

You can make a system of equations using this information.

$\left\{ {\begin{array}{*{20}{c}} {10e + 20c = 278}\\ {20e + 30c = 473} \end{array}} \right.$

Step 1: Write both equations in standard form, $Ax + By = C$ .

This system already is in standard form.

Step 2: Make the coefficients of one variable opposites.

$- 2\left( {10e + 20c} \right) = - 2\left( {278} \right)$

$- 20e - 40c = - 556$

Step 3: Add the equations resulting from Step 2 to eliminate one variable.

	–20e	–40c	=	–556
	20e	30c	=	473
SUM:		–10c	=	–83

Step 4: Solve for the remaining variable.

$- 10c = - 83$

$c = 8.3$

Step 5: Substitute the solution from Step 4 into one of the original equations. Solve for the remaining variable.

$10e + 20c = 278$

$10e + 20\left( {8.3} \right) = 278$

$10e + 166 = 278$

$10e = 112$

$e = 11.2$

Step 6: Interpret the answer.

Jenna burns 8.3 calories per minute circuit training and 11.2 calories per minute on the elliptical trainer.

Step 6: Checking is left to you.

Jenna burns 8.3 calories per minute circuit training and 11.2 calories per minute while on the elliptical trainer.

5.89

1.

$\left\{ {\begin{array}{*{20}{c}} {x - 5y > 10}\\ {2x + 3y > - 2} \end{array}} \right.$
$x - 5y > 10$	Substitute in the first inequality.
$3 - 5\left( { - 1} \right) > 10$
$3 + 6 > 10$	Simplify.
$9 > 10$

This is not true. This is not a solution to the system.

not a solution

2.

$\left\{ {\begin{array}{*{20}{c}} {x - 5y > 10}\\ {2x + 3y > - 2} \end{array}} \right.$
$x - 5y > 10$	Substitute in the first inequality.
$6 - 5\left( { - 3} \right) > 10$
$6 + 15 > 10$	Simplify.
$23 > 10$	This is true.
$2x + 3y > - 2$	Substitute in the second inequality.
$2\left( 6 \right) + 3\left( { - 3} \right) > - 2$
$12 - 9 > - 2$	Simplify.
$3 > - 2$

This is true. The ordered pair made both inequalities true. This is a solution to the system.

a solution

5.90

1.

Look at the graph. Find the region that satisfies both inequalities. To do this, pick a test point in the area shaded by both graphs and check.

Try (1, 7).
$\left\{ {\begin{array}{*{20}{c}} {y < 3x + 2}\\ {y > - x - 1} \end{array}} \right.$
$y < 3x + 2$	Substitute in the first inequality.
$7 < 3\left( 7 \right) + 2$
$7 < 21 + 2$	Simplify.
$7 < 23$	This is true.
$y > - x - 1$	Substitute in the second inequality.
$7 > - 1 - 1$
$7 > - 2$	Simplify. This is also true.

The ordered pair made both inequalities true. This is a solution to the system.

The region containing (0, 0) is the solution to the system of linear inequalities.

5.91

1.

$x + y \le 2$

Graph the first line as a solid line using intercepts: $\left( {2,0} \right)$ and $\left( {0,2} \right)$ .

Test (0, 0).
$0 + 0 \le 2$	This is true. Shade lightly the side where (0, 0) is located.
$y \ge \frac{2}{3}x - 1$

Graph the second line as a solid line using the intercept $\left( {0, - 1} \right)$ and the slope of $\frac{2}{3}$ .

Test (0, 0).
$0 \ge \frac{2}{3}\left( 0 \right) - 1$
$0 \ge - 1$	This is also true. Shade lightly the side where (0, 0) is located.

The solution is the region shaded by both graphs.

The solution is the darkest shaded region.

Two lines are plotted on a coordinate plane. The horizontal and vertical axes range from negative 10 to 10, in increments of 1. The first line passes through the points, (negative 7, 9), (0, 2), (2, 0), and (9, negative 7). The region below the line is shaded in dark blue. The second line passes through the points, (negative 9, negative 7), (0, negative 1), and (9, 5). The region above the line is shaded in light blue. The two lines intersect approximately at (1.8, 0.3). The region to the left of the intersection point and within the lines is shaded in gray.

5.92

1.

$y \ge 3x - 2$

Graph the first line as a solid line using the intercept $\left( {0, - 2} \right)$ and the slope of $3$ .

Test (0, 0).
$0 \ge 3\left( 0 \right) - 2$
$0 \ge - 2$	This is true. Shade lightly the side where (0, 0) is located.
$y < - 1$

Graph the second line as a dotted line using the intercept $\left( {0, - 1} \right)$ and the slope of $0$ . (It is a horizontal line).

Test (0, 0).
$0 < - 1$	This is not true. Shade lightly the side where (0, 0) is not located.

The solution is the region shaded by both graphs.

The solution is the lighter shaded region.

Two lines are plotted on a coordinate plane. The horizontal and vertical axes range from negative 10 to 10, in increments of 1. The first (dashed) line is horizontal and it passes through y equals negative 1. The region below the line is shaded in gray. The second (solid) line passes through the points, (negative 2, negative 8), (0, negative 2), and (3, 7). The region to the left of the line is shaded in dark blue. The two lines intersect approximately at (0.5, 1). The region below the intersection point and within the lines is shaded in light blue.

5.93

1.

$3x - 2y \ge 12$

Graph the first line as a solid line using the intercepts $\left( {4,0} \right)$ and $\left( {0, - 6} \right)$ .

Test (0, 0).
$\left( 0 \right)x - 2\left( 0 \right) \ge 12$
$0 \ge 12$	This is false. Shade lightly the side where (0, 0) is not located.
$y \ge \frac{3}{2}x + 1$

Graph the second line as a solid line using the intercept $\left( {0,1} \right)$ and the slope of $\frac{3}{2}$ .

Test (0, 0).
$0 \ge \frac{3}{2}\left( 0 \right) + 1$
$0 \ge 1$	This is not true. Shade lightly the side where (0, 0) is not located.

There is no overlap in this graph, no shared points There is no solution.

No solution.

5.94

1.

$y \ge 3x + 1$

Graph the first line as a solid line using the intercept $\left( {0,1} \right)$ and the slope of $3$ .

Test (0, 0).
$0 \ge 3\left( 0 \right) + 1$
$0 \ge 1$	This is false. Shade lightly the side where (0, 0) is not located.
$- 3x + y \ge - 4$
$y \ge 3x - 4$	Rearrange the inequality to avoid fractional intercepts.

Graph the second line as a solid line using the intercept $\left( {0, - 4} \right)$ and the slope of $3$ .

Test (0, 0).
$0 \ge 3\left( 0 \right) - 4$
$0 \ge - 4$	This is true. Shade lightly the side where (0, 0) is located.

The solution is the region shaded by both graphs.

The solution is the lighter shaded region.

5.95

1.

Let h = the number of hamburgers eaten.

Let c = the number of cookies eaten.

A hamburger has 240 calories and costs $1.40.

A cookie has 160 calories and costs $0.50.

Omar wants to spend less than or equal to $5.

Omar needs to eat 800 calories.

The number of hamburgers and hot dogs cannot be less than zero.

Make a table of what you know.

Hamburger calories	Cookie Calories	at least	800 calories
240h	160h	≥	800

Hamburger cost in $	Cookie cost in $	no more than	$5
1.40h	0.50c	≤	5

You can make a system of equations using this information.

$\left\{ {\begin{array}{*{20}{c}} {240h + 160c \ge 800}\\ \begin{array}{l} 1.4h + 0.5c \le 5\\ h \ge 0\\ c \ge 0 \end{array} \end{array}} \right.$

\left\{ \begin{array}{l}240h + 160C \ge 800\\1.40h + 0.50C \le 5\\h \ge 0\\C \ge 0\end{array} \right.

2.

$\left\{ {\begin{array}{*{20}{c}} {240h + 160c \ge 800}\\ \begin{array}{l} 1.4h + 0.5c \le 5\\ h \ge 0\\ c \ge 0 \end{array} \end{array}} \right.$

To avoid too much clutter on the graph, for the last two inequalities, just remember that the solution is only in quadrant I.

$240h + 160c \ge 800$	To make this easier to work with, divide both sides by 80.
$3h + 2c \ge 10$
$2c \ge - 3h + 10$	Solve for c.
$\frac{{2c}}{2} \ge \frac{{ - 3h + 10}}{2}$
$c \ge - \frac{3}{2}h + 5$
Test (0, 0).
$0 \ge - \frac{3}{2}\left( 0 \right) + 5$
$0 \ge 5$	This is false. Shade lightly the side where (0, 0) is not.

Graph this boundary with a solid line, an intercept at $\left( {0,5} \right)$ and a slope of $- \frac{3}{2}$ .

$1.4h + 0.5c \le 5$	Multiply by 2 to clear the fraction for what takes the place of “y.”
$2\left( {1.4h + 0.5c} \right) \le 2\left( 5 \right)$
$2.8h + c \le 10$
$c \le - 2.8h + 10$

Graph this boundary with a solid line, an intercept at $\left( {0,10} \right)$ , and a slope of $- 2.8$ .

Test (0, 0).
$1.4\left( 0 \right) + 0.5\left( 0 \right) \le 5$
$0 \le 5$	This is true. Shade lightly the side where (0, 0) is located.

The shaded area is between the two lines. This is the solution to the system.

Two lines are plotted on a coordinate plane. The horizontal and vertical axes range from negative 0 to 10, in increments of 1. The first line passes through the points, (0, 10), (2, 5), and (4, 0). The region above the line is shaded in light gray. The second line passes through the points, (0, 6), (2, 3), and (4, 0). The region below the line is shaded in dark blue. The region within the lines is shaded light blue.

3.

Test (3, 2).

$240h + 160c \ge 800$

$240\left( 3 \right) + 160\left( 2 \right) \ge 800$

$720 + 360 \ge 800$

1,080 \ge 800

This is true. Test the second inequality.

$1.4h + 0.5c \le 5$

$1.4\left( 3 \right) + 0.5\left( 2 \right) \le 5$

$4.2 + 1 \le 5$

5.2 \le 5

This is false. The ordered pair is not a solution to the system.

He cannot eat 3 hamburgers and 2 cookies because they cost more than $5.00.

The point

(3, 2)

is not in the solution region. Omar would not choose to eat 3 hamburgers and 2 cookies.

4.

Test (2, 4).

$240h + 160c \ge 800$

$240\left( 2 \right) + 160\left( 4 \right) \ge 800$

$480 + 640 \ge 800$

1,120 \ge 800

This is true. Test the second inequality.

$1.4h + 0.5c \le 5$

$1.4\left( 2 \right) + 0.5\left( 4 \right) \le 5$

$2.8 + 2 \le 5$

4.8 \le 5

This is true. The ordered pair is a solution to the system.

Omar can eat 2 hamburgers and 4 cookies. They provide enough calories and cost less than $5.

The point

(2, 4)

is in the solution region. Omar might choose to eat 2 hamburgers and 4 cookies.

5.96

1.

Let a = the number of bags of apples.

Let b = the number of bunches of bananas.

The profit on apples is $4 per bag and the profit on bananas is $6 per bunch.

Break the words into components.

The total profit	is	$4 times the number of bags of apples	plus	$6 times the number of bunches of bananas.
P	=	4a	+	6b

$P = 4a + 6b$

With

a =

the number of bags of apples sold, and

b =

the number of bunches of bananas sold, the objective function is

P = 4a + 6b

.

5.97

1.

Let x = the number of widgets.

Let y = the number of wadgets

It takes 20 minutes to make a widget and 28 minutes to make a wadget.

Break the words into components.

The total time	is	the time to make widgets	plus	the time to make wadgets
T	=	20x	+	28y

$T = 20x + 28y$

T = 20x + 28y

5.98

1.

Let a = the number of bags of apples.

Let b = the number of bunches of bananas.

The profit on apples is $4 per bag and the profit on bananas is $6 per bunch.

Constraints:

No more than a total of 20 bags and bunches on school grounds each day.
The maximum weight capacity of the fruit is 70 pounds.
A bag of apples weighs 3 pounds.
A bunch of bananas weighs 5 pounds.

Break the words into components.

The total profit	is	$4 times the number of bags of apples	plus	$6 times the number of bunches of bananas.
P	=	4a	+	6b

$P = 4a + 6b$

The number of bags of apples	plus	The number of bunches of bananas	can be no more than	20
a	+	b	≤	20

$a + b \le 20$

The number of bags of apples times 3 pounds	plus	The number of bunches of bananas times 5 pounds	can be no more than	70 pounds
3a	+	5b	≤	70

$3a + 5b \le 70$

The summary of this system:

Objective function: $P = 4a + 6b$

Constraints:

$a + b \le 20$

$3a + 5b \le 70$

The constraints are

a + b \leq 20

and

3a + 5b \leq 70

. The summary is:

P = 4a + 6b

,

a + b \leq 20

, and

3a + 5b \leq 70

.

5.99

1.

The constraints are:

15 \leq x \leq 22

13 \leq y \leq 19

So the system is:

T = 20x + 28y

15 \leq x \leq 22

13 \leq y \leq 19

5.100

1.

The maximum value for the profit

P

occurs when

x = 15

and

y = 5

. This means that to maximize their profit, the Robotics Club should sell 15 bags of apples and 5 bunches of bananas every day.

Let a = the number of bags of apples.

Let b = the number of bunches of bananas.

The profit on apples is $4 per bag and the profit on bananas is $6 per bunch.

Constraints:

No more than a total of 20 bags and bunches on school grounds each day.
The maximum weight capacity of the fruit is 70 pounds.
A bag of apples weighs 3 pounds.
A bunch of bananas weighs 5 pounds.

Break the words into components.

The total profit	is	$4 times the number of bags of apples	plus	$6 times the number of bunches of bananas.
P	=	4a	+	6b

$P = 4a + 6b$

The number of bags of apples	plus	The number of bunches of bananas	can be no more than	20
a	+	b	≤	20

$a + b \le 20$

The number of bags of apples times 3 pounds	plus	The number of bunches of bananas times 5 pounds	can be no more than	70 pounds
3a	+	5b	≤	70

3a + 5b \le 70

The summary of this system:

$P = 4a + 6b$

$a + b \le 20$

$3a + 5b \le 70$

Solve the system.

$P = 4a + 6b$

Assume the normal x-axis is the a-axis and the y-axis is the b-axis.

The maximum occurs at one of the corner points.

Graph the system.

$a + b \le 20$

The boundary of this graph intersects the axes at $\left( {20,0} \right)$ and $\left( {0,20} \right)$ .

Test (0,0).

0 + 0 \le 20

This is in the solution, so shade under the boundary in quadrant I.

Graph $3a + 5b \le 70$ .

The boundary of this graph intersects the axes at $\left( {23.\overline 3 ,0} \right)$ and $\left( {0,14} \right)$ .

Test (0,0).

0 + 0 \le 70

This is in the solution, so shade under the boundary in quadrant I.

The solution is the area shaded by both graphs. The corners of the shaded area are the intersection of the two boundary lines and the points $\left( {0,14} \right)$ , and $\left( {20,0} \right)$ . The maximum and minimum will occur at one of the corner points. You do not need to test (0,0) because you know the profit is 0 there.

Where do the boundaries of these inequalities cross?

$a + b \le 20$

$3a + 5b \le 70$

Solve $a + b = 20$ for a.

$a = 20 - b$

Substitute into the second boundary equations and solve for b.

$3\left( {20 - b} \right) + 5b = 70$

$60 - 3b + 5b = 70$

$2b = 10$

$b = 5$

Substitute into the first equation to find b.

$a + 5 = 20$

$a = 15$

Interpret.

The corner points of the solution are the intersection of the two boundaries and the two points: $\left( {0,14} \right)$ , and $\left( {20,0} \right)$ .

$\left( {0,14} \right)$ What is the profit? $P = 4a + 6b = 4(0) + 6(14) = \$ 84$

$\left( {20,0} \right)$ What is the profit? $P = 4a + 6b = 4(20) + 6(0) = \$ 80$

$\left( {5,15} \right)$ What is the profit? $P = 4a + 6b = 4(15) + 6(5) = 60 + 30 = \$ 90$

The maximum profit is made when selling 15 bags of apples and 5 bunches of bananas each day.

Check Your Understanding

1.

Juliette is 2 inches taller than Vivian.

You can test these with sample numbers. If Vivian is 50 inches tall, then Juliette is 52 inches tall.

$J = V + 2$
$52 = 50 + 2$	This works, so $J = V + 2$ works.
$V = J - 2$
$50 = 52 - 2$	This works, so $V = J - 2$ works.
$J + 2 = V$
$52 + 2 = 50$	This is false, so this does not work.
$J = V - 2$
$52 = 50 - 2$	This is false, so this does not work.

J = V + 2

,

V = J - 2

2.

Expressions relate numbers and variables to each other without involving an equal sign. Equations tell you that two quantities are equal. Look for the equal sign to tell you the difference. These are expressions:

$5x + 8$

$2n + 3m$

5x+8

,

2n+3m

3.

$\left( {8x + 12} \right) \div 4x - 2x$	Divide the first two terms by 4x.
$\left( {8x \div 4x} \right) + \left( {12 \div 4x} \right) - 2x$	Simplify.
$2 + 3x - 2x$

Like terms are constants or have the same variable and exponent. Add the coefficients and keep the same variable.

$2 + x$	This is not equal to 10x. Try again.
$8x + \left( {12 \div 4x} \right) - 2x$	Divide 12 by 4x.
$8x + \left( {3x} \right) - 2x$	Combine the like terms.
$9x$	This is not equal to 10. Try again.
$8x + 12x \div \left( {4x - 2x} \right)$	Subtract inside the parentheses.
$8x + 12x \div \left( {2x} \right)$	Divide by 2x.
$8x + 6$	This is not equal to 10. Try again.
$\left( {8x + 12x} \right) \div \left( {4x - 2x} \right)$	Do the operations inside the parentheses.
$\left( {20x} \right) \div \left( {2x} \right)$	Divide.
10	This is equal to 10.

The expression $\left( {8x + 12x} \right) \div \left( {4x - 2x} \right)$ equals 10.

(8x + 12x) \div \left( {4x - 2x} \right)

4.

$3{x^2} - 7x + 2$	Try each of the given numbers to see which results in 50.
$3{\left( { - 2} \right)^2} - 7\left( { - 2} \right) + 2$
$3\left( 4 \right) + 14 + 2 = 12 + 14 + 2 = 28$	That is not 50. Try again.
$3{\left( { - 3} \right)^2} - 7\left( { - 3} \right) + 2$
$3\left( 9 \right) + 21 + 2 = 27 + 23 = 50$	This worked.

The value of x that makes the result 50 is –3.

- 3

5.

$\left( {x - y} \right)\left( {x - y} \right)$	Use the distributive property.
$x\left( {x - y} \right) - y\left( {x - y} \right)$	Use the distributive property again.
$x\left( x \right) - x\left( y \right) - y\left( x \right) - y\left( { - y} \right)$	Simplify.
${x^2} - xy - xy + {y^2}$
${x^2} - 2xy + {y^2}$

Two of the answers only have two terms, which is clearly wrong.

One of the answers forgot that when you multiply a negative by a negative, the answer is a positive. It has the wrong sign on the squared y term.

The correct answer is ${x^2} - 2xy + {y^2}$ .

{x^2}-2xy + {y^2}

6.

3x\left( {3{x^2} + x-2} \right)

Using the distributive property “in reverse” you can factor out 3x.

$9{x^3} + 3{x^2} - 6x = 3x\left( {3{x^2} + x - 2} \right)$

7.

You can tell because you need multiplication to be able to generate an ${x^2}$ term by multiplication.

multiplication

8.

You can tell because you need division to be able to generate a $\frac{1}{x}$ term.

division

9.
It is a correct solution strategy.
Let
$\begin{array}{rcl}{x}&{ = }&{38}\\{8\left( {38-2} \right)}&{ \mathop = \limits^? }&{6\left( {38 + 10} \right)}\\{8\left( {36} \right)}&{ \mathop = \limits^? }&{6\left( {48} \right)}\\{288}&{ = }&{288 ✓}\\\end{array}$

10.
It is a correct solution strategy.
Let
$\begin{array}{rcl}{x}&{ = }&{- \,2}\\{7 + 4\left( {2 + 5\left( { - \,2} \right)} \right)}&{ \mathop = \limits^? }&{3\left( {6\left( { -\,2} \right) + 7} \right)-\left( {13\left( { -\,2} \right) + 36} \right)}\\{7 + 4\left( {2-10} \right)}&{ \mathop = \limits^? }&{3\left( { - 12 + 7} \right)-\left( { -\,26 + 36} \right) }\\{7 + 4\left( { -\,8} \right)}&{ \mathop = \limits^? }&{3\left( { - 5} \right)-\left( {10} \right) }\\{7-32}&{ \mathop = \limits^? }&{ -\,15-10}\\{-\,25}&{ = }&{-\,25✓}\end{array}$

11.
This is not a correct solution strategy. The negative sign is not distributed correctly in the second line of the solution strategy. The second line should read $8x + 7-2x + 9 = 22-4x + 4$ .

12.

Break your words into components.

The total fare is a flat fee of $3.00 plus $1.70 per mile

T = 3 + 1.7x

$T = 3 + 1.7x$

Alternate answer option: $T = 1.7x + 3$

${{T}} = 1.7x + 3$

13.

Substitute 22 for x in the equation $T = 3+1.7x$ .

$T = 3 + 1.7(22) = 3 + 37.4 = 40.4$

The ride from the airport to the hotel costs $40.40.

$40.40

14.

Break your words into components.

The total fare is a flat fee of $5.00 plus $1.60 per mile

T = 5 + 1.6y

$T = 5 + 1.6y$

Alternate answer option: $T = 1.6y + 5$

${{T}} = 1.6y + 5$

15.

Substitute 22 for y in the equation $T = 5 + 1.6y$ .

$T = 5 + 1.6(22) = 5 + 35.2 = 40.2$

The ride from the airport to the hotel costs $40.20.

$40.20

16.

The Enjoyable Cab Company costs $40.20, while the Nice Cab Company costs 20 cents more at $40.40. You should use the Enjoyable Cab Company.

$40.40 – 40.20 = 0.20$

The Enjoyable Cab Company, because the cab fare will be $ 0.20 less than what it would cost to take a taxi from the Nice Cab Company.

17.

$3\left( {2x - 3} \right) = 12\left( {x - 3} \right) - 3\left( {2x - 9} \right)$

Step 1: Simplify each side as much as possible.

$6x - 9 = 12x - 36 - 6x + 27$

$6x - 9 = 6x - 9$

Step 2: Collect the variable terms on one side.

$6x - 9 - 6x = 6x - 9 - 6x$

$- 9 = - 9$

This statement is always true. The variable does not matter at all, so there are infinitely many solutions.

Luis is right.

Luis is; there are infinitely many solutions. If this is solved using the general strategy, it simplifies to $0 = 0$ . This is a true statement, so there are infinitely many solutions.

18.
d $F = \frac{9}{5}C + 32$

$C = \frac{5}{9}\left( {F - 32} \right)$ Treat F as the variable and everything else as “constants.”

$\left( {\frac{9}{5}} \right)C = \left( {\frac{9}{5}} \right)\frac{5}{9}\left( {F - 32} \right)$ Multiply both sides by $\left( {\frac{9}{5}} \right)$ .

$\left( {\frac{9}{5}} \right)C = F - 32$

$\left( {\frac{9}{5}} \right)C + 32 = F - 32 + 32$ Add 32 to both sides.

$\frac{9}{5}C + 32 = F$ Switch sides to make it look like the answer options.

$F = \frac{9}{5}C + 32$ This is option (d).

19.
b
Break your words into components.

A temperature in Kelvin is the Celsius temperature plus 273 degrees

K = C + 273

$K = C + 273$

$K = C + 273$ , although d $C = K – 273$ is an equivalent formula.

20.
a
Use $C = \frac{5}{9}\left( {F - 32} \right)$ and K = C + 273 to find a Kelvin temperature given degrees Fahrenheit.

$K = C + {\rm{ }}273\;$

$K = \frac{5}{9}\left( {F - 32} \right) + 273$ Substitute the right side of the Fahrenheit to Celsius formula.

$K = \frac{5}{9}\left( {F-32} \right) + 273$

21.
b
Break your words into components.

A temperature in Rankin is the Fahrenheit temperature plus 460 degrees

R = F + 460

Use $F = \frac{9}{5}C + 32$ and $R = F + {\rm{ }}460\;$ to find a Rankin temperature given degrees Celsius.

$R = F + {\rm{ }}460\;$

$R = \frac{9}{5}C + 32 + 460$ Substitute the right side of the Fahrenheit to Celsius formula.

$R = \frac{9}{5}C + 492$

$R = \frac{9}{5}C + 492$

22.

The parenthesis indicates that 1 is not included in the solution. The region shaded is to the left of 1, so that means the solution includes numbers less than 1. In interval notation, this looks like $\left( { - \infty , - 1} \right)$ .

Notice that the interval notation version is in the same order as the graph. Negative infinity is on the left and 1 is on the right.

d $\left( { - \infty , - 1} \right)$

23.

The bracket indicates that 5 is included in the solution. The region shaded is to right of 5, so that means the solution includes numbers greater than 5. In interval notation, this looks like $[5,\infty )$ .

Notice that the interval notation version is in the same order as the graph. The number 5 is on the left and positive infinity is on the right.

b $[ - 5,\infty )$

24.

The bracket indicates that $\frac{3}{2}$ is included in the solution. The region shaded is to right of $\frac{3}{2}$ , so that means the solution includes numbers greater than $\frac{3}{2}$ . In interval notation, this looks like $\left[ {\frac{3}{2},\infty } \right)$ .

Notice that the interval notation version is in the same order as the graph. The number $\frac{3}{2}$ is on the left and positive infinity is on the right.

c $\left[ {\frac{3}{2},\infty } \right)$

25.

The solution is shaded between parentheses at –4 and 3. This corresponds in interval notation to those numbers between parentheses with a comma in between.

$\left( { - 4,3} \right)$

a $( - 4,3)$

26.
e

$6x \ge 24$ This is the correct answer.

$\frac{{6x}}{6} \ge \frac{{24}}{6}$ Divide by 6.

$x \ge 4$

$\left[ {4,\infty } \right)$ This is interval notation.

$6x \geq 24$

27.
b

$- 6x \ge 18$ This is the correct answer.

$\frac{{ - 6x}}{{ - 6}} \le \frac{{18}}{{ - 6}}$ Divide by –6 and switch the sign because you divided by a negative number.

$x \le - 3$

$\left( {-\infty ,-3} \right)$ This is interval notation.

$-6x \gt 18$

28.
c

$4x - 3 > - 11$ This is the correct answer.

$4x - 3 + 3 > - 11 + 3$ Add 3.

$4x > - 8$

$\frac{{4x}}{4} > \frac{{ - 8}}{4}$ Divide by 4.

$x > - 2$

$\left( {-2,\infty } \right)$ This is interval notation.

$4x - 3 > - 11$

29.
c

$- 3x + 14 > - 13$ This is the correct answer.

$- 3x + 14 - 14 > - 13 - 14$ Subtract 14

$- 3x > - 27$

$\frac{{ - 3x}}{{ - 3}} < \frac{{ - 27}}{{ - 3}}$ Divide by –3 and switch the sign since you divided by a negative number.

$x < 9$

$\left( {-\infty ,9} \right)$ This is interval notation.

$- 3x + 14 > - 13$

30.
b
Break your words into components.

Let x = the number of lawn chairs.

8 cubic feet = size of one box.

764 cubic feet = capacity of truck.

Break the words into components.

764 is the maximum for 8 times number of boxes

764 < 8 ∙ x

Solve the inequality $764 > 8x$

$764 > 8x$ Reverse to match one of the options.

$8x < 764$ This is correct. You do not need to solve further for this exercise.

$8x < 764$

31.
d
Break your words into components.

Let x = the number of nights to babysit.

$50 = the average amount earned per night babysitting.

$8,120 = the amount needed to buy a car.

Break the words into components.

$8,120 is the minimum needed 50 times number of nights babysitting

8,120 < 50 ∙ x

Solve the inequality $8,120 \ge 50x$

$8,120 < 50x$ Reverse to match one of the options.

$50x > 8,120$ This is correct. You do not need to solve further for this exercise.

$50x > \text{8,120}$

32.
a This is true, because the second proportions are the reciprocals of the first set of ratios. The qualification that all the numbers are nonzero prevents the problem of dividing by zero.
True

33.
a
$\frac{{wolves}}{{rabbits}} = \frac{3}{5}$

$\frac{{rabbits}}{{wolves + rabbits}} = \frac{5}{{3 + 5}} = \frac{5}{8}$

True

34.
a Ratios make part to part, part to whole, and whole to part comparisons. Fractions make part to whole comparisons only.
True

35.
b Cross multiplication cannot be used as the first step in this equation. You can only have a single rational expression on each side when you cross multiply. You would need to add the expressions on the right side so you had a single rational expression on the right.
False

36.
a Ratios make part to part, part to whole, and whole to part comparisons. Fractions make part to whole comparisons only.
True

37.
Since the words “math majors” come before the words “non-math majors,” the number of math majors needs to come before the number of non-math majors. The ratio 12:16 has the numbers reversed, with the non-math majors first.
$12{:}16$

38.

There are 16 math majors and 12 non-math majors.

There are 12 + 16 = 28 total students.

The ratio of math majors to all students is 16 to 28 or 16:28.

$16{:}28$

39.

Set up the two ratios as proportions.

Let x = the number of British pounds.

It helps many students to create a table to help set up their ratios.

Exchange What I know or want to know

U.S. Dollars 1 $450

British pounds 0.72 x

Set up your proportion.

$\frac{1}{{0.72}} = \frac{{450}}{x}$

$x = \left( {0.72} \right)\left( {450} \right)$ Cross multiply.

$x = 324$

Damon will get 324 British pounds in return, which is “None of these” in terms of the available answer options.

324.00 British pounds (None of these.)

40.
10 inches

Set up the two ratios as proportions.

Let x = the length of the model locomotive, in inches

It helps many students to create a table to help set up their ratios.

Scale What I know or want to know

Model train 1 x

Real train 87 73 feet = 876 inches

Set up your proportion.

$\frac{1}{{87}} = \frac{x}{{876}}$

$87x = \left( 1 \right)\left( {876} \right)$ Cross multiply.

$\frac{{87x}}{{87}} = \frac{{876}}{{87}}$ Divide.

$x \approx 10{\rm{\ inches}}$

The model locomotive is approximately 10 inches long.

41.

Set up the two ratios as proportions.

Let x = the distance Albert can go on one tank of gas.

It helps many students to create a table to help set up their ratios.

Milage of Car On one tank of gas

Miles 37 x

Gallon 1 13.5

Set up your proportion.

$\frac{{37}}{1} = \frac{x}{{13.5}}$

$x = \left( {37} \right)\left( {13.5} \right)$ Cross multiply.

$x = 499.5{\rm{ miles}}$

Yes he can, but barely. At 37 miles per gallon, Albert can drive $499.5$ miles. While in theory he can make it, he probably should fill up with gasoline somewhere along the way!

42.

Set up the two ratios as proportions.

Let x = the amount paid for the gas.

It helps many students to create a table to help set up their ratios.

Cost of one gallon The large transaction

Price $4.28 x

Gallons 1 9.5

Set up your proportion.

$\frac{{4.28}}{1} = \frac{x}{{9.5}}$

$x = \left( {4.28} \right)\left( {9.5} \right)$ Cross multiply.

$x = \$ 40.66$

Albert paid $40.66 for the 9.5 gallons of gas.

$$40.66$

43.
d
Test all the options by substitution until you find the option that works.

$6y + 10 = 12y$

$6\left( {\frac{5}{3}} \right) + 10 = 12\left( {\frac{5}{3}} \right)$

$10 + 10 = 20$

Since this is true, $\frac{5}{3}$ is a solution.

$y = \frac{5}{3}$

44.

Graph the equation and then compare it to the options offered.

$y = 3x + 5$

Find three points that are solutions to the equation. Choose three values for x.

x y

–1

0

1

Substitute into the equation to find the values for y.

x y $y=3x+5$

–1 2 $3(-1) + 5= -3 + 5 = 2$

0 5 $3(0) + 5 = 5$

1 8 $3(1) + 5 = 8$

Plot the points. Check that they line up and draw the line.

The first graph is the correct graph. It is the only one that passes through $(0, 5)$ .

45.
b
Test the points for all the options. Only the correct option works.

Correct option: $y = \frac{1}{2}x + 4$ .

$(0, 4)$

$4 = \frac{1}{2}\left( 0 \right) + 4$

4 = 4 is true, so this ordered pair is on the line.

$(-8, 0)$

$0 = \frac{1}{2}\left( { - 8} \right) + 4$

$0 = -4 + 4$

0 = 0 is true, so this ordered pair is on the line.

Two points are all it takes to determine a line, so this is the correct equation for the line.

$y = \frac{1}{2}x + 4$

46.

Graph the inequality, and then compare it to the options.

$y > 3x + 5$

Graph the line $y = 3x + 5$ with a dotted line, since this is a strictly greater than inequality.

Shade above the line since this is a “y >”.

Test $(0, 0)$ to be sure you shaded correctly.

$0 > 3\left( 0 \right) + 5$

$0 > 5$ is false, so this ordered pair is not on the line. $(0, 0)$ should not be in the shaded area. You shaded correctly.

Two points are all it takes to determine a line, so this is the correct equation for the line.

The correct graph is the third graph.

47.
b
Test the points on the line and test points in the shaded are for all the options. Only the correct option works.

Correct option: $y \le - 2x + 5$ .

$(0, 5)$

$5 \le - 2\left( 0 \right) + 5$

$5 \le 5$ is true, so this ordered pair is a solution to the inequality and a point on the boundary.

$(4, -3)$

$- 3 \le - 2\left( 4 \right) + 5$

$- 3 \le - 8 + 5$

$- 3 \le - 3$ is true, so this ordered pair is a solution to the inequality and a point on the boundary.

$(0, 0)$ to test shaded area.

$0 \le - 2\left( 0 \right) + 5$

$0 \le 5$ is true, so shade the side of the boundary where $(0, 0)$ is.

$y \leq -2x + 5$

48.
d

$\left({x - 3} \right)\left({x + 5} \right)$

$x\left({x + 5} \right) - 3\left({x + 5} \right)$ Distribute.

${x^2} + 5x - 3x - 15$ Distribute again.

${x^2} + 2x - 15$ Combine like terms.

This matches the correct option.

49.
b
${x^2} - 8x + 15$

Look for two factors of 15 that add up to –8.

Factors of 8 Sum of Factors = –8 Success?

1 times 15 $1 + 15 = 16$ No

3 times 5 $3 + 5 = 8$ No, but the sum is 8, so change signs.

–3 times –5 $-3 + \left({-5} \right) = -8$ Yes!

$\left({x - 3} \right)\left({x - 5} \right)$ Write the factored form.

This is the correct option.

50.
c
Since it intersects the $x$ -axis at –5 and –1, you can “reverse” solve.

$x = - 5$ $x = - 1$ Write the factors they came from.

$x + 5 = 0$ $x + 1 = 0$ Multiply together to get the quadratic equation.

$\left({x + 5} \right)\left({x + 1} \right) = 0$ Distribute twice to expand the polynomial.

${x^2} + 5x + x + 5 = 0$ Simplify.

${x^2} + 6x + 5 = 0$ This is the correct option.

51.
d

${x^2} = 49$

$x \pm \sqrt {49}$ Use the Square Root Property.

$x = \pm 7$ Simplify.

$x = 7$ $x = - 7$ Rewrite the two solutions.

This is the correct option.

52.
b This is false. When you distribute in the product, you get $-x$ and $-5x.$ This adds up to $-6x,$ which is not the middle term of the trinomial.
False

53.
b False. You cannot use the square root method when you have a linear term. The $5x$ term prevents you from using the square root method unless you do a lot of rearranging this equation.
False

54.
a True. Any quadratic equation can be solved with the quadratic formula. Some will have imaginary solutions, so you are not guaranteed a real number solution.
True

55.
b
False. For example, substitute $(1, 0)$ into the equation.

Substitute 1 in for $x .$ You expect the result to be 0, but it is 1.

${(1)^2} - 5(1) + 5 = 1 - 5 + 5 = 1$

False

56.

${x^2} - 5x + 5 = 0$

This is the completing the square method. First, you must make the left side look like an expression squared. It is usually a good method to choose when the linear term has an even coefficient, but it can be done here.

Move the constant to the right side.

${x^2} - 5x = - 5$

Let $b =$ the linear term’s coefficient, so $b = -5$

Add ${\left({\frac{b}{2}} \right)^2}$ to both sides. ${\left({\frac{b}{2}} \right)^2} = {\left({\frac{{ - 5}}{2}} \right)^2} = \frac{{25}}{4}$

${x^2} - 5x + \frac{{25}}{4} = - 5 + \frac{{25}}{4}$ Now, you can write the left side as an expression squared.

${\left({x - \frac{5}{2}} \right)^2} = - \frac{{20}}{4} + \frac{{25}}{4}$ Add the constants on the right.

${\left({x - \frac{5}{2}} \right)^2} = \frac{5}{4}$ You can use the Square Root Property in this form.

$x - \frac{5}{2} = \pm \sqrt {\frac{5}{4}}$

$x - \frac{5}{2} = \pm \frac{{\sqrt 5 }}{2}$ Simplify.

$x = \frac{5}{2} \pm \frac{{\sqrt 5 }}{2}$

$x = \frac{5}{2} + \frac{{\sqrt 5 }}{2}$ $x = \frac{5}{2} - \frac{{\sqrt 5 }}{2}$ Write the two solutions.

Checking is left for you.

$x = \frac{5}{2} \pm \frac{{\sqrt 5 }}{2}$

57.
a
$f\left( x \right) = 2x - 8$

$f\left( 3 \right) = 2\left( 3 \right) - 8$

$f\left( 3 \right) = 6 - 8$

$f\left( 3 \right) = - 2$

True

58.
b This relation is not a function because there are repeated x-values. The x-value 3 is matched with more than one y-value.
False

59.
b This is not a function. There are places where a vertical line can pass through it more than once. It fails the vertical line test.
False

60.
a True
This relation is a function because there are no repeated x-values. Each x-value is matched with only one y-value.

61.
b The domain is the set of all x-values of the relation. {10, 20, 30, 40}
False

62.
a
Substitute 0 for y to find the x-intercept.

$y = 2(0) - 8 = - 8$

(0, –8) is the x-intercept.

True

63.
b
The slope is 2, not 1.

Point 1: (1, 2) and Point 2: (2, 4).

Find the slope.

$m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{4 - 2}}{{2 - 1}}$ Substitute.

$m = \frac{2}{1} = 2$ Simplify.

The slope is 2.

You can plot the points and verify the slope by counting the rise over the run.

False

64.
b Count the rise over run between the two labeled points. The rise is 4. The run is 1. The rise over run is 4, not 5.
False

65.
a
The slope is 4. The y-intercept is (0, 5).

When the equation is in the slope-intercept form of the equation of the line,

$y = mx + b$ .

m = the slope.

(0, b) = the y-intercept.

This line is $y = 4x + 5$ .

True

66.
b Vertical lines have an undefined slope. This is also expressed as no slope.
False

67.
Elimination is best because you can multiply the second equation by 5 to eliminate the y variable.
Elimination

68.
Substitution is a good choice because the first equation is already solved for x.
Substitution

69.
Substitution is a good choice because the first equation is already solved for y.
Substitution

70.
Elimination is best because you can multiply the second equation by –4 to eliminate the x variable.
Elimination

71.
Substitution is a good choice because the first equation is already solved for y.
Substitution

72.
Elimination is best because you can multiply the second equation by –3 to eliminate the x variable.
Elimination

73.
Elimination is best because you can multiply the second equation by –2 to eliminate the x variable.
Elimination

74.
Substitution is a good choice because the first equation is already solved for x.
Substitution

75.
c

76.
e
Graph the system, then identify which graph matches what you drew.

$2x - 3y \le 5$ Graph the first line as a solid line using the intercepts $\left( {\frac{5}{2},0} \right)$ and $\left( {0, - \frac{5}{3}} \right)$ .

Test (0, 0).

$2\left( 0 \right) + 3\left( 0 \right) \le 5$

$0 \le 5$ This is true. Shade lightly the side where (0, 0) is located.

Second line: $x - y \ge 3$ .

Graph the second line as a solid line using the intercepts $\left( {3,0} \right)$ and $\left( {0, - 3} \right)$ .

Test (0, 0).

$0 - 0 \ge 3$ This is false. Shade lightly the side where (0, 0) is not located.

The solution is the region shaded by both graphs.

77.
d
Graph the system, then identify which graph matches what you drew.

$2x - 3y > 5$ Graph the first line as a dotted line using the intercepts $\left( {\frac{5}{2},0} \right)$ and $\left( {0, - \frac{5}{3}} \right)$ .

Test (0, 0).

$2\left( 0 \right) - 3\left( 0 \right) > 5$

$0 > 5$ This is false. Shade lightly the opposite side of where (0, 0) is located.

Second line: $x + y < 3$ .

Graph the second line as a dotted line using the intercepts $\left( {3,0} \right)$ and $\left( {0,3} \right)$ .

Test (0, 0).

$0 - 0 < 3$ This is true. Shade lightly the side where (0, 0) is located.

The solution is the region shaded by both graphs.

78.
b
Graph the system, then identify which graph matches what you drew.

$2x + 3y > 5$ Graph the first line as a dotted line using the intercepts $\left( {\frac{5}{2},0} \right)$ and $\left( {0,\frac{5}{3}} \right)$ .

Test (0, 0).

$2\left( 0 \right) + 3\left( 0 \right) > 5$

$0 > 5$ This is false. Shade lightly the opposite side of where (0, 0) is located.

Second line: $x + y \le 3$ .

Graph the second line as a solid line using the intercepts $\left( {3,0} \right)$ and $\left( {0,3} \right)$ .

Test (0, 0).

$0 - 0 \le 3$ This is true. Shade lightly the side where (0, 0) is located.

The solution is the region shaded by both graphs.

79.
a
Graph the system, then identify which graph matches what you drew.

$2x + 3y \le 5$ Graph the first line as a solid line using the intercepts $\left( {\frac{5}{2},0} \right)$ and $\left( {0,\frac{5}{3}} \right)$ .

Test (0, 0).

$2\left( 0 \right) + 3\left( 0 \right) \le 5$

$0 \le 5$ This is true. Shade lightly the side where (0, 0) is located.

Second line: $x + y \ge 3$ .

Graph the second line as a solid line using the intercepts $\left( {3,0} \right)$ and $\left( {0,3} \right)$ .

Test (0, 0).

$0 + 0 \ge 3$ This is false. Shade lightly the side where (0, 0) is not located.

The solution is the region shaded by both graphs.

80.
c
Let t = the number of tables sold.

Let c = the number of chairs sold.

Let P = the profit.

The profit on a table is $20 and the profit on a chair is $10.

Break the words into components.

The total profit is $20 times the number of tables sold plus $10 times the number of chairs sold

P = 20t + 10c

$P = 20t + 10c$

81.
a
Let w = the number of acres of wheat.

Let c = the number of acres of barley.

Let P = the profit.

The profit on an acre of wheat is $150 and the profit on an acre of barley is $180.

Break the words into components.

The total profit is $150 times the number of acres of wheat plus $180 times the number of acres of barley

P = 150w + 180b

$P = 150w + 180b$

82.
b
Let f = the number of 45 rpm records.

Let t = the number of 33 rpm records.

Let P = the profit.

The profit on a 45-rpm record is $2.50.

The profit on a 33-rpm record is $6.75.

Break the words into components.

The total profit is $2.50 times the number of 45-rpm records sold plus $6.75 times the number of 33-rpm records sold

P = 2.50f + 6.75t

$P = 2.50f + 6.75t$

83.
d
Let t = the number of tables sold.

Let c = the number of chairs sold.

Let P = the profit.

The profit on a table is $20 and the profit on a chair is $10.

A table requires 15 board feet of wood.

A chair requires 4 board feet of wood.

Kellie has 70 board feet available.

Constraint:

Break the words into components.

15 board feet times the number of tables plus 4 board feet times the number of chairs must be no more than 70 board feet

15t + 4c ≤ 70

$15t + 4c \le 70$

84.
a
Let t = the number of tables sold.

Let c = the number of chairs sold.

Let P = the profit.

The maximum number of tables and chairs made in a day is 12.

Break the words into components.

The number of tables plus the number of chairs is no more than 12

t + c ≤ 12

$t + c \le 12$

85.
c
Let w = the number of acres of wheat.

Let b = the number of acres of barley.

The cost of seed is $10 per acre for wheat.

The cost of seed is $15 per acre for barley.

Dave can spend only $945 on seed.

Constraint:

Break the words into components.

Cost of wheat seed times number of acres plus Cost of barley seed times number of acres is no more than $945

10w + 15b ≤ 945

$10w + 15b \le 945$

86.
d
Let w = the number of acres of wheat.

Let b = the number of acres of barley.

The profit on an acre of wheat is $150 while the cost is $30 per acre.

The profit on an acre of barley is $180 while the cost is $25 per acre.

Dave budgets $1,635 for raising both crops.

Constraint:

Break the words into components.

Cost of wheat crop times number of acres plus Cost of barley crop times number of acres is no more than $1,635

30w + 25b ≤ 1,635

$30w + 25b \le 1,635$

87.

Let t = the number of tables sold.

Let c = the number of chairs sold.

Let P = the profit.

The profit on a table is $20 and the profit on a chair is $10.

A table requires 15 board feet of wood.

A chair requires 4 board feet of wood.

Kellie has 70 board feet available.

The maximum number of tables and chairs made in a day is 12.

Constraint:

Break the words into components.

15 board feet times the number of tables plus 4 board feet times the number of chairs must be no more than 70 board feet

15t + 4c ≤ 70

$15t + 4c \le 70$

The number of tables plus the number of chairs must be no more than 12

t + c ≤ 12

$t + c \le 12$

Consider t to be the “x-axis” variable and c to be the “y-axis” variable.

$15t + 4c \le 70$

Graph the boundary with a solid line using the intercepts $\left( {12,0} \right)$ and $\left( {0,12} \right)$ .

Test (0,0).

$0 + 0 \le 12$ This is true, so shade below the boundary line in quadrant I.

Graph $15t + 4c \le 70$ .

Graph the boundary with a solid line using the intercepts $\left( {4.\overline 6 ,0} \right)$ and $\left( {0,17.5} \right)$ .

Test (0,0).

$0 + 0 \le 70$ This is true, so shade below the boundary line in quadrant I.

The solution is the area shaded by both graphs.

88.
b
Where do the boundaries of these inequalities cross?

$t + c \le 12$

$15t + 4c \le 70$

Solve $t + c = 12$ for t.

$t = 12 - c$

Substitute into the second boundary equations and solve for b.

$15\left( {12 - c} \right) + 4c = 70$

$180 - 15c + 4c = 70$

$- 11c = - 110$

$c = 10$

Substitute into the first equation to find b.

$t = 12 - 10$

$t = 2$

The corners of the shaded area are

the intersection of the two boundary lines: (2, 10)

the points where two of the boundary lines intersect the axes: $\left( {0,12} \right)$ , and $\left( {4.\frac{2}{3},0} \right)$

(0, 0).

89.

Let P = the profit.

The profit on a table is ${\text{\$20}}$ and the profit on a chair is ${\text{\$10}}$ .

$P = 20t + 10c$

The maximum and minimum will occur at one of the corner points. You do not need to test (0,0) because you know the profit is 0 there, even though technically it is also a corner point.

Interpret.

The corner points of the solution are the intersection of the two boundaries, (2, 10), and the two points: $\left( {0,12} \right)$ , and $\left( {4.\frac{2}{3},0} \right)$ .

$\left( {0,12} \right)$ What is the profit? $P = 20\left( 0 \right) + 10\left( {12} \right) = \$120$ .

$\left( {4.\frac{2}{3},0} \right)$ What is the profit? $P = 20\left( {\frac{{14}}{3}} \right) + 10\left( 0 \right) \approx \$93.33$ .

$\left( {2,10} \right)$ What is the profit? $P = 20\left( 2 \right) + 10\left( {10} \right) = \$140$ .

The maximum profit is made when making 2 tables and 10 chairs. Kellie will make a profit of ${\text{\$140}}$ .

$$140$

The total fare	is	a flat fee of $3.00	plus	$1.70 per mile
T	=	3	+	1.7x

The total fare	is	a flat fee of $5.00	plus	$1.60 per mile
T	=	5	+	1.6y

$C = \frac{5}{9}\left( {F - 32} \right)$	Treat F as the variable and everything else as “constants.”
$\left( {\frac{9}{5}} \right)C = \left( {\frac{9}{5}} \right)\frac{5}{9}\left( {F - 32} \right)$	Multiply both sides by $\left( {\frac{9}{5}} \right)$ .
$\left( {\frac{9}{5}} \right)C = F - 32$
$\left( {\frac{9}{5}} \right)C + 32 = F - 32 + 32$	Add 32 to both sides.
$\frac{9}{5}C + 32 = F$	Switch sides to make it look like the answer options.
$F = \frac{9}{5}C + 32$	This is option (d).

A temperature in Kelvin	is	the Celsius temperature	plus	273 degrees
K	=	C	+	273

A temperature in Rankin	is	the Fahrenheit temperature	plus	460 degrees
R	=	F	+	460

$R = F + {\rm{ }}460\;$
$R = \frac{9}{5}C + 32 + 460$	Substitute the right side of the Fahrenheit to Celsius formula.
$R = \frac{9}{5}C + 492$

$6x \ge 24$	This is the correct answer.
$\frac{{6x}}{6} \ge \frac{{24}}{6}$	Divide by 6.
$x \ge 4$
$\left[ {4,\infty } \right)$	This is interval notation.

$- 6x \ge 18$	This is the correct answer.
$\frac{{ - 6x}}{{ - 6}} \le \frac{{18}}{{ - 6}}$	Divide by –6 and switch the sign because you divided by a negative number.
$x \le - 3$
$\left( {-\infty ,-3} \right)$	This is interval notation.

$4x - 3 > - 11$	This is the correct answer.
$4x - 3 + 3 > - 11 + 3$	Add 3.
$4x > - 8$
$\frac{{4x}}{4} > \frac{{ - 8}}{4}$	Divide by 4.
$x > - 2$
$\left( {-2,\infty } \right)$	This is interval notation.

$- 3x + 14 > - 13$	This is the correct answer.
$- 3x + 14 - 14 > - 13 - 14$	Subtract 14
$- 3x > - 27$
$\frac{{ - 3x}}{{ - 3}} < \frac{{ - 27}}{{ - 3}}$	Divide by –3 and switch the sign since you divided by a negative number.
$x < 9$
$\left( {-\infty ,9} \right)$	This is interval notation.

$764 > 8x$	Reverse to match one of the options.
$8x < 764$	This is correct. You do not need to solve further for this exercise.

$8,120	is the minimum needed	50	times	number of nights babysitting
8,120	<	50	∙	x

$8,120 < 50x$	Reverse to match one of the options.
$50x > 8,120$	This is correct. You do not need to solve further for this exercise.

$\frac{1}{{0.72}} = \frac{{450}}{x}$
$x = \left( {0.72} \right)\left( {450} \right)$	Cross multiply.
$x = 324$

	Scale	What I know or want to know
Model train	1	x
Real train	87	73 feet = 876 inches

$\frac{1}{{87}} = \frac{x}{{876}}$
$87x = \left( 1 \right)\left( {876} \right)$	Cross multiply.
$\frac{{87x}}{{87}} = \frac{{876}}{{87}}$	Divide.
$x \approx 10{\rm{\ inches}}$

$\frac{{37}}{1} = \frac{x}{{13.5}}$
$x = \left( {37} \right)\left( {13.5} \right)$	Cross multiply.
$x = 499.5{\rm{ miles}}$

	Cost of one gallon	The large transaction
Price	$4.28	x
Gallons	1	9.5

$\frac{{4.28}}{1} = \frac{x}{{9.5}}$
$x = \left( {4.28} \right)\left( {9.5} \right)$	Cross multiply.
$x = \$ 40.66$

x	y	$y=3x+5$
–1	2	$3(-1) + 5= -3 + 5 = 2$
0	5	$3(0) + 5 = 5$
1	8	$3(1) + 5 = 8$

Chapter 5

Your Turn

Check Your Understanding

$\left({x - 3} \right)\left({x + 5} \right)$
$x\left({x + 5} \right) - 3\left({x + 5} \right)$	Distribute.
${x^2} + 5x - 3x - 15$	Distribute again.
${x^2} + 2x - 15$	Combine like terms.

Factors of 8	Sum of Factors = –8	Success?
1 times 15	$1 + 15 = 16$	No
3 times 5	$3 + 5 = 8$	No, but the sum is 8, so change signs.
–3 times –5	$-3 + \left({-5} \right) = -8$	Yes!

$x = - 5$	$x = - 1$	Write the factors they came from.
$x + 5 = 0$	$x + 1 = 0$	Multiply together to get the quadratic equation.
$\left({x + 5} \right)\left({x + 1} \right) = 0$		Distribute twice to expand the polynomial.
${x^2} + 5x + x + 5 = 0$		Simplify.
${x^2} + 6x + 5 = 0$		This is the correct option.

${x^2} = 49$
$x \pm \sqrt {49}$		Use the Square Root Property.
$x = \pm 7$		Simplify.
$x = 7$	$x = - 7$	Rewrite the two solutions.

${x^2} - 5x + \frac{{25}}{4} = - 5 + \frac{{25}}{4}$		Now, you can write the left side as an expression squared.
${\left({x - \frac{5}{2}} \right)^2} = - \frac{{20}}{4} + \frac{{25}}{4}$		Add the constants on the right.
${\left({x - \frac{5}{2}} \right)^2} = \frac{5}{4}$		You can use the Square Root Property in this form.
$x - \frac{5}{2} = \pm \sqrt {\frac{5}{4}}$
$x - \frac{5}{2} = \pm \frac{{\sqrt 5 }}{2}$		Simplify.
$x = \frac{5}{2} \pm \frac{{\sqrt 5 }}{2}$
$x = \frac{5}{2} + \frac{{\sqrt 5 }}{2}$	$x = \frac{5}{2} - \frac{{\sqrt 5 }}{2}$	Write the two solutions.

$m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{4 - 2}}{{2 - 1}}$	Substitute.
$m = \frac{2}{1} = 2$	Simplify.

The total profit	is	$20 times the number of tables sold	plus	$10 times the number of chairs sold
P	=	20t	+	10c

The total profit	is	$150 times the number of acres of wheat	plus	$180 times the number of acres of barley
P	=	150w	+	180b

The total profit	is	$2.50 times the number of 45-rpm records sold	plus	$6.75 times the number of 33-rpm records sold
P	=	2.50f	+	6.75t

15 board feet times the number of tables	plus	4 board feet times the number of chairs	must be no more than	70 board feet
15t	+	4c	≤	70