Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 5 Answer Key Writing Linear Equations.

## Texas Go Math Grade 8 Module 5 Answer Key Writing Linear Equations

**Essential Question**

How can you use linear equations to solve real world problems?

**Texas Go Math Grade 8 Module 5 Are You Ready? Answer Key**

**Complete these exercises to review skills you will need for this chapter.**

**Write each fraction as a decimal.**

Question 1.

\(\frac{1}{8}\) ________

Answer:

Given,

\(\frac{1}{8}\)

Convert from fraction to the decimal.

= 1 ÷ 8

= 0.125

Question 2.

\(\frac{0.3}{0.5}\) ________

Answer:

Given,

\(\frac{0.3}{0.5}\)

Convert from fraction to the decimal.

\(\frac{3}{10}\) ÷ \(\frac{5}{10}\) = \(\frac{3}{5}\) = 0.6

Question 3.

\(\frac{0.13}{0.8}\) ________

Answer:

Given,

\(\frac{0.13}{0.8}\)

Convert from fraction to decimal.

\(\frac{13}{10}\) ÷ \(\frac{8}{10}\) = \(\frac{13}{8}\) = 1.625

Question 4.

\(\frac{0.39}{0.52}\) ________

Answer:

Given,

\(\frac{0.39}{0.52}\)

Convert from fraction to the decimal.

\(\frac{39}{100}\) ÷ \(\frac{52}{100}\) = \(\frac{39}{52}\) = 0.75

**Solve each equation using the inverse operation.**

Question 5.

7p = 28 __________________

Answer:

7p = 28 p is multiplied by 7

\(\frac{7 p}{7}\) = \(\frac{28}{7}\) To solve the equation, use the inverse operation, division.

p = 4

Question 6.

h – 13 = 5 ________

Answer:

h – 3 = 13 is subtracted from h

h – 13 + 13 = 5 + 13 To solve the equation, use the inverse operation, addition.

h = 18

Question 7.

\(\frac{y}{3}\) = -6 ______

Answer:

\(\frac{y}{3}\) = -6 y is divided by 3

\(\frac{y}{3}\) * 3 = -6 * 3 To solve the equation, use the inverse operation, multiplication.

y = -18

Question 8.

b + 9 = 21 _________

Answer:

b + 9 = 21 9 is added to b

b + 9 – 9 = To solve the equation, use the inverse operation, subtraction.

b = 12

Question 9.

c – 8 = -8 ________________

Answer:

c – 8 = 8 8 is subtracted from c

c – 8 + 8 = -8 + 8 To solve the equation, use the inverse operation, addition

c = 0

Question 10.

3n = -12 _______

Answer:

3n = -12

n is multiplied by 3

\(\frac{3 n}{3}\) = \(\frac{-12}{3}\) To solve the equation, use the inverse operation, division.

n = -4

Question 11.

-16 = m + 7 _______

Answer:

-16 = m + 7 7 is added to m

-16 – 7 = m + 7 To solve the equation, use the inverse operation, subtraction.

m = -23

Question 12.

\(\frac{t}{-5}\) = -5 ____

Answer:

\(\frac{t}{-5}\) = -5

As we can see t is divided by (-5), so to solve the equation, we use the inverse operation: muLtiplication.

\(\frac{t}{-5}\) . (-5) = (-5) . (-5)

t = 25

**Texas Go Math Grade 8 Module 5 Reading Start-Up Answer Key**

**Visualize Vocabulary**

**Use the ✓ words to complete the diagram. You can put more than one word in each bubble.**

**Understand Vocabulary**

**Complete the sentences using the preview words.**

Question 1.

A set of data that is made up of two paired variables is ____

Answer:

A set of data that is made up of two paired variables is __Bivariate data
__Bivariate data is defined as a set of data that is made up of two o paired variables. Each value of one of the variables is paired with a value of the other variable.

By definition, a set of data that is made up of two variables is bivariate data.

Thus, the blank part of the statement is bivariate data.

Question 2.

When the rate of change varies from point to point, the relationship is a ____.

Answer:

When the rate of change varies from point to point, the relationship is a __nonlinear
__A nonlinear relationship is a type of relationship between two points in which change in one point does not correspond with change in other point. It is shown when the rate of change varies between pairs of points.

By description, when the rate of change varies from point to point the relationship is nonlinear.

Thus, the blank part of the statement is nonlinear.