**Algebra 1 Common Core Answers Student Edition Grade 8 – 9 Chapter 1 Foundations for Algebra Exercise 1.8**

Algebra 1 Common Core Answers Student Edition Grade 8 – 9

**Chapter 1 Foundations for Algebra Exercise 1.8 1CB**

The amount y (in ton) of waste placed in the landfill in x months is given by the equation

y = 560x

The aim is to find the number of months required to accumulate 11,200 t of waste in the landfill by making a table on a graphical calculator.

We need to find the value of x when y = 11,200.

On a T-83 Plus graphical calculator,

The value of x for which y = 11,200 is x = 20.

It will take 20 months to accumulate 11,200 t of waste in the landfill.

**Chapter 1 Foundations for Algebra Exercise 1.8 1LC**

Consider the equation

y + 1 = 8

The aim is to check whether y = -9 a solution of the equation y + 1 = 8.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

First substitutein the equation and then check whether it is true.

**Chapter 1 Foundations for Algebra Exercise 1.8 2CB**

The amount y (in dollars) of the customer’s purchase, if x is the amount (in dollars) of the purchase before the coupon is used, is given by the equation

y = x – 15

The aim is to find the original price of a shirt if a customer using the coupon pays $17.

We need to find the value of x when y = 17.

Prepare a table of values using a graphical calculator.

On a T-83 Plus graphical calculator,

The value of x for which y = 17 is x = 32.

Therefore, the original price of the shirt was $32.

**Chapter 1 Foundations for Algebra Exercise 1.8 2LC**

Consider the equation

x – 3 = 12

The aim is to find the solution of the equation using mental math.

The phrase that describes the equation is “a number minus 3 equals 12”.

Ask yourself “What number minus 3 is equal to 12?”

The answer is 15.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

**Chapter 1 Foundations for Algebra Exercise 1.8 3LC**

**Chapter 1 Foundations for Algebra Exercise 1.8 4LC**

**Chapter 1 Foundations for Algebra Exercise 1.8 5LC**

An equation is an open sentence if it contains one or more variables and may be true or false depending on the values of its variables.

The aim is to write an open equation containing one variable and division.

An example of an open equation containing one variable(x) and division is

**Chapter 1 Foundations for Algebra Exercise 1.8 6LC**

**Chapter 1 Foundations for Algebra Exercise 1.8 7E**

Consider the equation 85 + (-10) = 95

The aim is to test whether the equation is true, false, or open.

Remember that:

- An equation is true if the expressions on either side of the equal sign are equal.
- An equation is false if the expressions on either side of the equal sign are not equal.
- An equation is an open sentence if it contains one or more variables and may be true or false depending on the values of its variables.

Since the equation does not contain variable(s), it is not an open sentence.

Simplify each side of the equation and check whether they are equal.

Since the expressions on either side of the equal sign are not equal, the equation is false.

**Chapter 1 Foundations for Algebra Exercise 1.8 8E**

Consider the equation 225 ÷ t – 4 = 6.4

The aim is to test whether the equation is true, false, or open.

Remember that:

- An equation is true if the expressions on either side of the equal sign are equal.
- An equation is false if the expressions on either side of the equal sign are not equal.
- An equation is an open sentence if it contains one or more variables and may be true or false depending on the values of its variables.

Since the equation contains a variable, t, it is an open sentence.

**Chapter 1 Foundations for Algebra Exercise 1.8 9E**

Consider the equation 29 – 34 = -5

The aim is to test whether the equation is true, false, or open.

Remember that:

- An equation is true if the expressions on either side of the equal sign are equal.
- An equation is false if the expressions on either side of the equal sign are not equal.
- An equation is an open sentence if it contains one or more variables and may be true or false depending on the values of its variables.

Since the equation does not contain variable(s), it is not an open sentence.

Simplify each side of the equation and check whether they are equal.

Since the expressions on either side of the equal sign are equal, the equation is true.

**Chapter 1 Foundations for Algebra Exercise 1.8 10E**

Consider the equation -8(-2) – 7 = 14 – 5

The aim is to test whether the equation is true, false, or open.

Remember that:

- An equation is true if the expressions on either side of the equal sign are equal.
- An equation is false if the expressions on either side of the equal sign are not equal.

Since the equation does not contain variable(s), it is not an open sentence.

Simplify each side of the equation and check whether they are equal.

Since the expressions on either side of the equal sign are equal, the equation is true.

**Chapter 1 Foundations for Algebra Exercise 1.8 11E**

Consider the equation 4(-4) ÷ (-8)6 = -3 + 5(3)

The aim is to test whether the equation is true, false, or open.

Remember that:

- An equation is true if the expressions on either side of the equal sign are equal.
- An equation is false if the expressions on either side of the equal sign are not equal.

Since the equation does not contain variable(s), it is not an open sentence.

Simplify each side of the equation and check whether they are equal.

Since the expressions on either side of the equal sign are not equal, the equation is false.

**Chapter 1 Foundations for Algebra Exercise 1.8 12E**

Consider the equation 91 ÷ (-7) – 5 = 35 ÷ 7 + 3

The aim is to test whether the equation is true, false, or open.

Remember that:

- An equation is true if the expressions on either side of the equal sign are equal.
- An equation is false if the expressions on either side of the equal sign are not equal.

Since the equation does not contain variable(s), it is not an open sentence.

Simplify each side of the equation and check whether they are equal.

Since the expressions on either side of the equal sign are not equal, the equation is false.

**Chapter 1 Foundations for Algebra Exercise 1.8 13E**

Consider the equation 4a – 3b = 21

The aim is to test whether the equation is true, false, or open.

Remember that:

- An equation is true if the expressions on either side of the equal sign are equal.
- An equation is false if the expressions on either side of the equal sign are not equal.

Since the equation contains variables a and b, it is an open sentence.

**Chapter 1 Foundations for Algebra Exercise 1.8 14E**

Consider the equation 14 + 7 + (-1) = 21

The aim is to test whether the equation is true, false, or open.

Remember that:

- An equation is true if the expressions on either side of the equal sign are equal.
- An equation is false if the expressions on either side of the equal sign are not equal.

Simplify each side of the equation and check whether they are equal.

Since the expressions on either side of the equal sign are not equal, the equation is false.

**Chapter 1 Foundations for Algebra Exercise 1.8 15E**

Consider the equation 5x + 7 = 17

The aim is to test whether the equation is true, false, or open.

Remember that:

- An equation is true if the expressions on either side of the equal sign are equal.
- An equation is false if the expressions on either side of the equal sign are not equal.

Since the equation contains a variable, x, it is an open sentence.

**Chapter 1 Foundations for Algebra Exercise 1.8 16E**

Consider the equation 8x + 5 = 29

The aim is to check whether x = 3 a solution of the equation.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

First substitute x = 3 in the equation and then check whether it is true.

**Chapter 1 Foundations for Algebra Exercise 1.8 17E**

Consider the equation 5b + 1 = 16

The aim is to check whether b = -3 a solution of the equation.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

First substitute b = -3 in the equation and then check whether it is true.

**Chapter 1 Foundations for Algebra Exercise 1.8 18E**

Consider the equation 6 = 2n – 8

The aim is to check whether n = 7 a solution of the equation.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

First substitute n = 7 in the equation and then check whether it is true.

**Chapter 1 Foundations for Algebra Exercise 1.8 19E**

Consider the equation 2 = 10 – 4y

The aim is to check whether y = 2 a solution of the equation.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

First substitute y = 2 in the equation and then check whether it is true.

**Chapter 1 Foundations for Algebra Exercise 1.8 20E**

Consider the equation 9a – (-72) = 0

The aim is to check whether a = -8 a solution of the equation.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

First substitute a = -8 in the equation and then check whether it is true.

**Chapter 1 Foundations for Algebra Exercise 1.8 21E**

**Chapter 1 Foundations for Algebra Exercise 1.8 22E**

**Chapter 1 Foundations for Algebra Exercise 1.8 23E**

**Chapter 1 Foundations for Algebra Exercise 1.8 24E**

**Chapter 1 Foundations for Algebra Exercise 1.8 25E**

**Chapter 1 Foundations for Algebra Exercise 1.8 26E**

**Chapter 1 Foundations for Algebra Exercise 1.8 27E**

The athlete trains for 115 min each day.

In d days, the athlete will be trains for 115 × d = 115d min.

If the athlete trains for 690 min in d days, 115d must be equal to 690.

115d = 690

Therefore, the equation that relates the number of days d that the athlete spends training when the athlete trains for 690 min is 115d = 690.

**Chapter 1 Foundations for Algebra Exercise 1.8 28E**

The manager of the restaurant earns $2.25 more each hour than the host of the restaurant.

If the host of the restaurant earns the amount h (in dollars) each hour, the manager will earn the amount

(h + 2.25)(in dollars), which must be equal to $11.50.

h + 2.25 = 11.50

Therefore, the equation that relates the amount h that the host of the restaurant earns each hour when the manager earns $11.50 each hour is h + 2.25 = 11.50.

**Chapter 1 Foundations for Algebra Exercise 1.8 29E**

Consider the equation x – 3 = 10

The aim is to find the solution of the equation using mental math.

The phrase that describes the equation is “a number minus 3 equals 10”.

Ask yourself “What number minus 3 is equal to 10?”

The answer is 13.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

Let us check whether x = 13 a solution of the equation x – 3 = 10.

First substitute x = 13 in the equation and then check whether it is true.

x – 3 = 10

x – 3 = 10 Substitute x = 13

10 = 10 Simplify each side

The simplified equation 10 = 10 is true, because each side of the equal sign is 10.

So, x = 13 is a solution of the equation x – 3 = 10.

Therefore, the solution of the equation is x = 13.

**Chapter 1 Foundations for Algebra Exercise 1.8 30E**

Consider the equation 4 = 7 – y

The aim is to find the solution of the equation using mental math.

The phrase that describes the equation is “4 is equal to 7 minus a number”.

Ask yourself “What number when subtracted from 7 gives 4?”

The answer is 3.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

Let us check whether y = 3 a solution of the equation 4 = 7 – y.

**Chapter 1 Foundations for Algebra Exercise 1.8 31E**

Consider the equation 18 + d = 24

The aim is to find the solution of the equation using mental math.

The phrase that describes the equation is “the sum of 18 and a number is 24”.

Ask yourself “What number when added to 18 gives 24?”

The answer is 6.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

**Chapter 1 Foundations for Algebra Exercise 1.8 32E**

**Chapter 1 Foundations for Algebra Exercise 1.8 33E**

**Chapter 1 Foundations for Algebra Exercise 1.8 34E**

**Chapter 1 Foundations for Algebra Exercise 1.8 35E**

**Chapter 1 Foundations for Algebra Exercise 1.8 36E**

Consider the equation 20a = 100

The aim is to find the solution of the equation using mental math.

The phrase that describes the equation is “a number multiplied by 20 is 100”.

Ask yourself “What number when multiplied by 20 gives 100?”

The answer is 5.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

**Chapter 1 Foundations for Algebra Exercise 1.8 37E**

Consider the equation 13c = 26

The aim is to find the solution of the equation using mental math.

The phrase that describes the equation is “a number multiplied by 13 is 26”.

Ask yourself “What number when multiplied by 13 gives 26?”

The answer is 2.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

**Chapter 1 Foundations for Algebra Exercise 1.8 38E**

**Chapter 1 Foundations for Algebra Exercise 1.8 39E**

**Chapter 1 Foundations for Algebra Exercise 1.8 40E**

**Chapter 1 Foundations for Algebra Exercise 1.8 41E**

**Chapter 1 Foundations for Algebra Exercise 1.8 42E**

**Chapter 1 Foundations for Algebra Exercise 1.8 43E**

**Chapter 1 Foundations for Algebra Exercise 1.8 44E**

**Chapter 1 Foundations for Algebra Exercise 1.8 45E**

**Chapter 1 Foundations for Algebra Exercise 1.8 46E**

**Chapter 1 Foundations for Algebra Exercise 1.8 47E**

**Chapter 1 Foundations for Algebra Exercise 1.8 48E**

**Chapter 1 Foundations for Algebra Exercise 1.8 49E**

**Chapter 1 Foundations for Algebra Exercise 1.8 50E**

**Chapter 1 Foundations for Algebra Exercise 1.8 51E**

An expression is a mathematical relationship between numbers and variables.

For example, 3x + 5 is an expression.

On the other side, an equation shows that two expressions are equal.

For example, 2x + 1 = 5 is an equation.

An expression does not have a solution, however, can be simplified.

**Chapter 1 Foundations for Algebra Exercise 1.8 52E**

The number of peoples attend a basketball team’s championship game is 1254.

The number of identical benches in the gymnasium is 6.

The aim is to find the number of peoples to seat in each bench, which tells us that we will be dividing.

We need to divide 1254 peoples in 6 benches.

Divide 1254 by 6.

1254 ÷ 6 = 209

Therefore, 209 peoples are expected to seat in each bench.

**Chapter 1 Foundations for Algebra Exercise 1.8 53E**

Consider the equation x + 4 = -2

The aim is to find (or estimate) the solution of the equation.

Use mental math.

The phrase that describes the equation is “the sum of a number and 4 is -2”.

Ask yourself “What number when added to 4 gives -2?”

The answer is -6.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

**Chapter 1 Foundations for Algebra Exercise 1.8 54E**

**Chapter 1 Foundations for Algebra Exercise 1.8 55E**

**Chapter 1 Foundations for Algebra Exercise 1.8 56E**

Consider the equation -3 + t = 19

The aim is to find (or estimate) the solution of the equation.

Use mental math.

The phrase that describes the equation is “the sum of -3 and a number is 19”.

Ask yourself “What number when added to -3 gives 19?”

The answer is 22.

Remember that a solution of an equation containing a variable is a value of the variable that makes the equation true.

**Chapter 1 Foundations for Algebra Exercise 1.8 57E**

**Chapter 1 Foundations for Algebra Exercise 1.8 58E**

**Chapter 1 Foundations for Algebra Exercise 1.8 59E**

**Chapter 1 Foundations for Algebra Exercise 1.8 60E**

**Chapter 1 Foundations for Algebra Exercise 1.8 61E**

**Chapter 1 Foundations for Algebra Exercise 1.8 62E**

The drill advances at the rate of 67 m/h.

The aim is to find the time will it take the drill to reach a depth of 300 m, which tells us that we will be dividing 300 by 67.

Divide 300 by 67.

300 ÷ 67 = 4.5 (Rounded to first decimal place)

Therefore, the drill will take 4.5 hours to reach the depth of 300 m.

To model the situation, first identify the variables.

Here the variables are depth and time.

Let us suppose that the drill takes t hours to reach a depth of d meter.

Since the drill advances at the rate of 67 m/h, it will reach at the depth of (67t) meter in t hours.

So, we must have

d = 67t

Therefore, the equation that model the situation is

d = 67t,

Where d is the depth (in meter) reach by the drill in t hours.

To model the situation, we need the rate 67 m/h at which the drill advances through the ice sheet.

**Chapter 1 Foundations for Algebra Exercise 1.8 63E**

**Chapter 1 Foundations for Algebra Exercise 1.8 64E**

**Chapter 1 Foundations for Algebra Exercise 1.8 65E**

Consider the equation 15 = 4 + 2t

The solution of the equation is the value of t for which the value of the expression on the right side of the equal sign will be 15.

Since 2 is an even number, for any integer t, 2t will be even number.

Since the sum or difference of two even numbers is an even number and 4 is an even number, ( 4 + 2t)will be an even number for any integer t.

The number on the right side of the equal sign is 15, which is an odd number.

So, the equation can’t have an integer solution, the solution must be a fraction.

A fraction must lie in between two consecutive integers.

So, the solution of the equation is between two consecutive integers.

**Chapter 1 Foundations for Algebra Exercise 1.8 66E**

The construction crew needs to install 550 ft of curbing.

Yesterday the crew installed 272 ft of curbing.

Today the crew needs to install ( 550 – 272) = 278 ft of curbing, if it wants to finish the job.

The crew can install curbing at the rate of 32 ft/h.

To find the time the crew will take to install 278 ft of curbing at the rate 32 ft/h, we will be dividing.

We need to divide 278 by 32.

Divide 278 by 32.

278 ÷ 32 = 8.7 (Rounded to one decimal place)

The crew will take 8.7 hours to install 278 ft of curbing.

**Chapter 1 Foundations for Algebra Exercise 1.8 67E**

**Chapter 1 Foundations for Algebra Exercise 1.8 68E**

**Chapter 1 Foundations for Algebra Exercise 1.8 69E**

**Chapter 1 Foundations for Algebra Exercise 1.8 70E**

**Chapter 1 Foundations for Algebra Exercise 1.8 71E**

**Chapter 1 Foundations for Algebra Exercise 1.8 72E**

**Chapter 1 Foundations for Algebra Exercise 1.8 73E**

**Chapter 1 Foundations for Algebra Exercise 1.8 74E**

**Chapter 1 Foundations for Algebra Exercise 1.8 75E**

**Chapter 1 Foundations for Algebra Exercise 1.8 76E**

**Chapter 1 Foundations for Algebra Exercise 1.8 77E**

**Chapter 1 Foundations for Algebra Exercise 1.8 78E**

**Chapter 1 Foundations for Algebra Exercise 1.8 79E**

**Chapter 1 Foundations for Algebra Exercise 1.8 80E**

**Chapter 1 Foundations for Algebra Exercise 1.8 81E**

**Chapter 1 Foundations for Algebra Exercise 1.8 82E**

**Chapter 1 Foundations for Algebra Exercise 1.8 83E**

**Chapter 1 Foundations for Algebra Exercise 1.8 84E**

**Chapter 1 Foundations for Algebra Exercise 1.8 85E**

**Chapter 1 Foundations for Algebra Exercise 1.8 86E**