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3.2: Solve Equations Involving Inverse Properties of Addition and Division

Difficulty Level: Basic Created by: CK-12
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Practice Two-Step Equations with Addition and Division
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Jessica and Casey worked at a bakery during school vacation. One day Casey was asked to divide up many pounds of flour. She divided the amount she was given by three. Then she added four more pounds to one of these portions.

Jessica was given the largest portion. If Jessica received 8 pounds of flour, how many pounds of flour did Casey begin with?

Do you know how to solve this problem? To figure it out, you will need to write and solve a two-step equation. Pay attention to this Concept and you will know how to solve it by the end of the Concept.

Guidance

You are going to learn how to solve two step equations. Let’s begin.

To solve a two-step equation, we will need to use more than one inverse operation. Let's take a look at how to solve a two-step equation now. When we perform inverse operations to find the value of a variable, we work to get the variable alone on one side of the equals. This is called isolating the variable. It is one strategy for solving equations. You can use isolating the variable whether you are solving one step or two step equations.

Solve for $c$ : $5 + \frac{c}{4} = 15$ .

Notice that there are two terms on the left side of the equation, 5 and $\frac{c}{4}$ .

Our first step should be to use inverse operations to get the term that includes a variable, $\frac{c}{4}$ , by itself on one side of the equal (=) sign.

In the equation, 5 is added to $\frac{c}{4}$ . So, we can use the inverse of addition—subtraction. We can subtract 5 from both sides of the equation.

$5 + \frac{c}{4} & = 15\\5 - 5 + \frac{c}{4} & = 15 - 5\\0 + \frac{c}{4} & = 10\\\frac{c}{4} & = 10$

Now, the term that includes a variable, $\frac{c}{4}$ , is by itself on one side of the equation.

We can now use inverse operations to get the $c$ by itself. Since $\frac{c}{4}$ means $c \div 4$ , we can use the inverse of division—multiplication. We can multiply both sides of the equation by 4.

$\frac{c}{4} & = 10\\\frac{c}{4} \times 4 & = 10 \times 4\\\frac{c}{4} \times \frac{4}{1} & = 40$

The number 4, or $\frac{4}{1}$ , is the multiplicative inverse , or reciprocal , of $\frac{1}{4}$ . You can find the multiplicative inverse of a number by flipping its numerator and its denominator. So, the multiplicative inverse of $\frac{4}{1}$ is $\frac{1}{4}$ . When a number is multiplied by its multiplicative inverse, the product is 1.

$\frac{c}{4} \times \frac{4}{1} & = 40\\c \times \left ( \frac{1}{4} \times \frac{4}{1} \right ) & = 40\\c \times 1 & = 40\\c & = 40$

The work above shows how multiplying each side of the equation by 4 isolates the variable.

Because 4 is the multiplicative inverse, or reciprocal, of $\frac{1}{4}$ , we could also have solved this problem by canceling out the 4's like this:

$\frac{c}{\bcancel{4}} \times \frac{\bcancel{4}}{1} & = 40\\\frac{c}{1} & = 40\\c & = 40$

The answer is that $c$ is equal to 40.

Let’s review our steps to solving this two step equation.

Take a few minutes to write these steps in your notebook.

Example A

$\frac{x}{5} + 6 = 10$

Solution: $x = 20$

Example B

$\frac{x}{9} + 12 = 28$

Solution:  $x = 144$

Example C

$\frac{x}{11} + 12 = 18$

Solution:  $x = 66$

Now let's go back to the dilemma from the beginning of the Concept.

Think about what we know. We know that Casey divided the pounds of flour by three, but we don't know how many pounds she started with, so this our variable.

$\frac{x}{3}$

Next, we know that Casey added four pounds to one of the portions.

$\frac{x}{3} + 4$

Jessica ended up with 8 pounds.

$\frac{x}{3} + 4 = 8$

Now we can solve the equation. Start by subtracting four from both sides of the equation.

$\frac{x}{3} + 4 - 4 = 8 - 4$

$\frac{x}{3} = 4$

Next we use the inverse of division, multiplication, and multiply three times four.

$x = 12$

Casey started with twelve pounds of flour.

Vocabulary

Equation
a mathematical statement with an equal sign where the quantity on one side of the equation is equal to the quantity on the other side.
Variable
a letter used to represent an unknown quantity.
Algebraic Equation
An equation with at least one variable in it.
One Step Equation
An algebraic equation with one operation in it.
Two Step Equation
An algebraic equation with two operations in it.

Guided Practice

Here is one for you to try on your own.

$\frac{y}{19} + 6 = 10$

Solution

First, we have to subtract 6 from each side of the equation.

$\frac{y}{19} = 10 - 6$

$\frac{y}{19} = 4$

Now we can multiply 19 times 4. This will give us the value of $y$ .

$(19)(4) = 76$

$y = 76$

This is our solution.

Practice

Directions: Solve the following two step equations that have addition and division in them.

1. $\frac{x}{3} + 4 = 8$
2. $\frac{x}{5} + 8 = 10$
3. $\frac{a}{6} + 7 = 13$
4. $\frac{a}{9} + 4 = 30$
5. $\frac{b}{8} + 6 = 15$
6. $\frac{c}{12} + 9 = 18$
7. $\frac{x}{7} + 7 = 21$
8. $\frac{x}{11} + 5 = 12$
9. $\frac{x}{12} + 9 = 16$
10. $\frac{a}{14} + 6 = 8$
11. $\frac{x}{22} + 9 = 12$
12. $\frac{y}{2} + 14 = 18$
13. $\frac{x}{7} + 24 = 38$
14. $\frac{x}{8} + 15 = 30$
15. $\frac{x}{9} + 11 = 28$

Vocabulary Language: English

Algebraic Equation

Algebraic Equation

An algebraic equation contains numbers, variables, operations, and an equals sign.
Equation

Equation

An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.
One-Step Equation

One-Step Equation

A one-step equation is an algebraic equation with one operation in it that requires one step to solve.
Two-Step Equation

Two-Step Equation

A two-step equation is an algebraic equation with two operations in it that requires two steps to solve.
Variable

Variable

A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.

Basic

Dec 19, 2012

Feb 26, 2015