Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 3 Lesson 3.3 Understanding Linear Equations with Two Variables to finish your assignments.

## Math in Focus Grade 7 Course 3 A Chapter 3 Lesson 3.3 Answer Key Understanding Linear Equations with Two Variables

### Math in Focus Grade 8 Chapter 3 Lesson 3.3 Guided Practice Answer Key

**Complete.**

Question 1.

Write a linear equation for the relationship between days, d, and weeks, w. A week has days. So a linear equation for d in terms of w is .

Answer:

Linear equation between days and weeek is

d= 7w,

Explanation:

Assuming there are 7 days in a week, W weeks will have 7 X W days,

So d = 7w.

Question 2.

George rented a car from a company for a week. The table shows the rental charges.

Answer:

A linear equation for C in terms of d is c = 100+(0.1 X d),

Explanation:

**Technology Activity**

Materials:

- graphing calculator

Use A Graphing Calculator To create Tables Of Values For Linear Equations With Two Variables

Work individually or in pairs.

You can use a graphing calculator to create tables of values for linear equations with two variables. Use the steps to create a table of values for the equation y = \(\frac{x}{\pi}\). This will give you the diameter y of a circle when you substitute a given circumference x.

Step 1.

Enter the equation using the equation screen.

Step 2.

Set the table function to use values of x starting at 0, with increments of 1.

Step 3.

Display the table. It will be in two columns as shown.

Step 4.

Repeat Step 1 to Step 2 for the equation y = -2x + \(\sqrt{2}\).

Find the value of y when x = -4

Answer:

9.414,

Explanantion:

When x =-4, y = -2x + \(\sqrt{2}\),

y = -2(-4) + 1.414

y= 8+ 1.414,

y = 9.414.

Question 3.

y = 7 + 3x

Answer:

y= -5,

Explanation:

Given y = 7 +3x,

y= 7 +3(-4),

y= 7 – 12,

y= -5.

Question 4.

\(\frac{1}{3} y\) = 2\(\left(x-\frac{1}{6}\right)\)

Answer:

y = -25

Explanation:

Question 5.

-6x – y = 17.75

Answer:

y = -41.75,

Explanation:

Given -6x – y = 17.75 when x =- 4

-6(-4)-y = 17.75,

-24 -17.75 = y

y = – 41.75.

Create a table of x- and y-values for each equation. Use integer values of x from 1 to 3.

Question 6.

2y = 1.2x + 1

Answer:

When x =1, y= 1.1, x= 2 , y = 1.7 and when x = 3, y is 2.3,

Explanation:

Question 7.

4y – 11x = 6

Answer:

When x =1, y = 17/4 or 4.25,

x= 2, y = 7 and x =3, y = 39/4 or 9.75,

Explanation:

**Complete the table of x- and y-values for each equation.**

Question 8.

Answer:

When y =-1 x is -1,

when x = 0 y is 2 and

when x =1 y is 5,

Explanation:

Question 9.

3(x + 1) – 2y = 0

Answer:

y = 9, x is 5, when y =16 1/2, x is 10 and

y = 24, x = 15,

Explanation:

### Math in Focus Course 3A Practice 3.3 Answer Key

**Write a linear equation for the relationship between the given quantities.**

Question 1.

meters, m, and centimeters, c

Answer:

1 m = 100 c

Explanation:

As 1 meter is equal to 100 centimeters so

1 m = 100 c.

Question 2.

hours, h, and seconds, s

Answer:

1 h= 3600 s,

Explanation:

As 1 hour is equal to 3600 seconds so

1h = 3600 s.

Question 3.

feet, f, and inches, i

Answer:

1 f= 12 i,

Explanation:

As 1 foot is equal to 12 inches so

1f = 12 i.

Question 4.

dollars, d, and cents, c

Answer:

1 d = 100 c

Explanation:

As 1 dollar is equal to 100 cents so

1 d = 100 c.

**Find the value of y when x = 2.**

Question 5.

2x – 1 = y + 4

Answer:

y = -1,

Explanation:

Given 2x -1 = y+4, substituting x =2 we get

2 X 2 -1 = y +4,

4 -1 = y +4,

3 = y + 4,

y = 3 – 4 = -1.

Question 6.

y = \(\frac{1}{7}\)(x + 5)

Answer:

y = 1,

Explanation:

Given y = \(\frac{1}{7}\)(x + 5) and x =2

y = \(\frac{1}{7}\)(2 + 5)= \(\frac{1}{7}\)(7) = 1.

Question 7.

3x – 11 = 2(y – 4)

Answer:

y = 3/2 or \(\frac{3}{2}\),

Explanation:

Given 3x – 11= 2(y-4),

3 X 2 – 11 = 2y – 8,

6 -11 +8 = 2y,

14 – 11 = 2y

3=2y,

y = 3/2.

Question 8.

4y = 5(x – 1)

Answer:

y = 5/4 or \(\frac{5}{4}\),

Explanation:

Given 4y=5(x-1) and x =2,

4y =5(2-1),

4y = 5(1),

4y = 5,

y = 5/4.

**Find the value of x when y = -7.**

Question 9.

2(3x – 7) = 9y

Answer:

x = -49/6 or –\(\frac{49}{6}\),

Explanation:

Given 2(3x-7) = 9y and y =-7,

6x – 14 = 9 X -7,

6x – 14 = -63,

6x = -63 +14

6x = -49

x= -49/6.

Question 10.

\(\frac{2 x-1}{5}\) = 2(y + 7)

Answer:

x = 1/2,

Explanation:

Given \(\frac{2 x-1}{5}\) = 2(y + 7) when y =-7,

2x -1 = 10(-7+7),

2x – 1 = 0,

2x = 1,

x= 1/2.

Question 11.

2x + y = 0.1(y + 3)

Answer:

x= 3.3,

Explanation:

Given 2x+y =0.1(y +3) when y=-7,

2x – 7 = 0.1(-7+3),

2x – 7 = 0.1(-4),

2x -7 = -0.4,

2x = -0.4 + 7,

2x = 6.6,

x= 6.6/2 = 3.3.

Question 12.

2y – 5x = 26

Answer:

y = -9/2 or \(\frac{-9}{2}\),

Explanation:

Given 2y – 5x = 26, when x = -7,

2y -5(-7)= 26,

2y + 35 = 26,

2y = 26-35,

2y=-9

y = -9/2.

**Create a table of x- and y-values for each equation. Use integer values of x from 1 to 3.**

Question 13.

y = \(\frac{1}{4}(8-x)\)

Answer:

x = 1, y = 7/4, x=2, y =3/2 and

when x = 3 y = 5/4,

Explanation:

Question 14.

x + 7 = \(\frac{1}{2}(y-5)\)

Answer:

x=1, y = 21,

x=2, y = 23,

x = 3, y =25,

Explanation:

Question 15.

-4y = 2x + 5

Answer:

x = 1, y = -7/4,

x = 2, y =-9/4,

x= 3, y = -11/4,

Explanation:

Question 16.

\(\frac{1}{2}(x+4)\) = \(\frac{1}{3}(y+1)\)

Answer:

x=1, y = 13/2,

x= 2, y =8,

x =3 then y = 19/2,

Explanation:

**Complete the table of x- and y-values for each equation.**

Question 17.

y = 5(x + 3)

Answer:

x= 0, y =15, x= 1, y = 20, x = 2 , y =25,

Explanation:

Question 18.

\(\frac{x}{4}\) + y = 1

Answer:

x = 2, y = 1/2 or 0.5,

y =0, x = 4 and

when y = -0.5, x = 6,

Explanation:

Question 19.

3x – 4y = \(\frac{5}{3}\)

Answer:

y =-2 2/3, x =-3,

x=-2, y =-23/12 and

x = -1 , y = -7/6,

Explanation:

Question 20.

5(y + 4) = 8x

Answer:

y = -4, x = 0, y = 12 , x = 10 and when y = 28 , x is 20,

Explanation:

Solve. Show your work.

Question 21.

A research student recorded the distance traveled by a car for every gallon of gasoline used.

She recorded the results in the following table.

Write a linear equation for the distance traveled,

d miles, in terms of the amount of gasoline used, g gallons.

Answer:

d=40.5g,

Explanation:

Given a research student recorded the distance traveled by a car for every gallon of gasoline used.

She recorded the results in the following table. For amount of Gasoline used (g gallons)

for 1 gallons it is 40.5 miles, for 2 gallons it is 2 X 40.5 = 81 miles,

for 3 gallons it is 3 X 40.5 = 121.5 miles and for 4 gallons it is 4 X 40.5 = 162 miles it is

linear so the equation is d = 40.5 g respectively.

Question 22.

Mr. Taransky sells blood pressure monitors. He earns a monthly salary

that includes a basic amount of $750 and $5 for each monitor sold.

a) Write a linear equation for his monthly salary, M dollars, in terms of the number of monitors sold, n.

Answer:

M= $750 +$5n,

Explanation:

Given Mr. Taransky sells blood pressure monitors. He earns a monthly salary

that includes a basic amount of $750 and $5 for each monitor sold.

Writing a linear equation for his monthly salary, M dollars, in terms of the number of monitors sold, n.

M = $750 +$5n.

b) Use the equation to complete the table of values below.

Answer:

Explanation:

Completed the table as shown above.

c) Calculate his salary when he sold 300 monitors in a month.

Answer:

Mr. Taransky salaray is $2,250 when he sold 300 monitors in a month,

Explanation:

Given to calculate his salary when he sold 300 monitors in a month,

M = $750 +$5 X 300,

M = $750 + $1,500 = $2,250.

d) His salary was $1,450 in January. How many monitors did he sell that month?

Answer:

140 monitors,

Explanation:

Given his salary was $1,450 in January, Monitors did he sell that month is

$1,450 = $750+$5n,

$5n = $1450-$750,

$5n = $700,

n = 700/5 = 140 monitors.

Question 23.

Bernadette’s cell phone plan costs a basic charge of $20 a month plus

5c per minute of calling time after she uses the first 300 minutes of calling time in a month.

a) Write a linear equation for the total cost, C dollars, in terms of the total

number of minutes of calling time, t minutes.

Answer:

Linear equation is c = 20 + 0.05(t -300),

Explanation:

Given Bernadette’s cell phone plan costs a basic charge of $20 a month plus

5c per minute of calling time after she uses the first 300 minutes of calling time in a month.

The linear equation for the total cost, C dollars, in terms of the total

number of minutes of calling time, t minutes is c = 20 + 0.05(t -300).

b) Create a table of t- and C-values for the linear equation. Use t = 300, 400, and 500.

Answer:

When t = 300, c = 20, t=400 c = 25 and when t = 500 c is 30,

Explanation:

c) Find the total number of minutes of calling time Bernardette uses in a month when her bill is $32.

Answer:

540 minutes,

Explanation:

As we have c =20+0.05(t-300) the total number of minutes of calling time

Bernardette uses in a month when her bill is $32

$32 = 20 +0.05(t-300),

$32 – $20 = 0.05 (t-300),

$12 = 0.05(t -300)

t – 300 = 12/0.05 = 240

t = 240 +300 = 540 minutes.

Question 24.

A parking lot charges $1.50 for the first hour or part of an hour. After the first hour, parking costs 70¢ for each half hour, or part of a half hour. Mr. Fischer parked his car in the parking lot for t hours.

a) Write a linear equation for the total cost of parking, y dollars, in terms of t.

Answer:

y= $1.50 + 2(0.7)(t-1)

Explanation:

Given a parking lot charges $1.50 for the first hour or part of an hour.

After the first hour, parking costs 70¢ for each half hour, or part of a half hour.

Mr. Fischer parked his car in the parking lot for t hours.

linear equation for the total cost of parking, y dollars, in terms of t.

y = $1.50 +2(0.7)(t-1).

b) Find the amount Mr. Fischer had to pay if he parked his car for a 40 minutes.

Answer:

Mr. Fischer has to pay $1.50,

Explanation:

Given the amount Mr. Fischer had to pay if he parked his car for a 40 minutes as

t hours is less than 1 hour, we take t as 1,

Mr.Fischer has to pay y= $1.50 +2(0.7)(1-1),

y= $1.50 + 0

y = $1.50.

c) What was the total cost if he parked for 1 hour and 40 minutes?

Answer:

The total cost if he parked for 1 hour and 40 minutes is $2.90,

Explanation:

Given the parked time is 1 hour and 40 minutes more than an

hour and half hour so we take t as 2,

y = $1.50+2(0.7)(2-1) = $1.50 + 2 X 0.7,

y= $1.50 +$1.40 = $2.90.