3.2: Solve Equations Involving Inverse Properties of Addition and Division
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 twostep 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 twostep equation, we will need to use more than one inverse operation. Let's take a look at how to solve a twostep 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 \begin{align*}c\end{align*}
Notice that there are two terms on the left side of the equation, 5 and \begin{align*}\frac{c}{4}\end{align*}
Our first step should be to use inverse operations to get the term that includes a variable, \begin{align*}\frac{c}{4}\end{align*}
In the equation, 5 is added to \begin{align*}\frac{c}{4}\end{align*}
\begin{align*}5 + \frac{c}{4} & = 15\\ 5  5 + \frac{c}{4} & = 15  5\\ 0 + \frac{c}{4} & = 10\\ \frac{c}{4} & = 10\end{align*}
Now, the term that includes a variable, \begin{align*}\frac{c}{4}\end{align*}
We can now use inverse operations to get the \begin{align*}c\end{align*}
\begin{align*}\frac{c}{4} & = 10\\ \frac{c}{4} \times 4 & = 10 \times 4\\ \frac{c}{4} \times \frac{4}{1} & = 40\end{align*}
The number 4, or \begin{align*}\frac{4}{1}\end{align*}
\begin{align*}\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\end{align*}
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 \begin{align*}\frac{1}{4}\end{align*}
\begin{align*}\frac{c}{\bcancel{4}} \times \frac{\bcancel{4}}{1} & = 40\\ \frac{c}{1} & = 40\\ c & = 40\end{align*}
The answer is that \begin{align*}c\end{align*}
Let’s review our steps to solving this two step equation.
Take a few minutes to write these steps in your notebook.
Example A
\begin{align*}\frac{x}{5} + 6 = 10\end{align*}
Solution: \begin{align*}x = 20\end{align*}
Example B
\begin{align*}\frac{x}{9} + 12 = 28\end{align*}
Solution: \begin{align*}x = 144\end{align*}
Example C
\begin{align*}\frac{x}{11} + 12 = 18\end{align*}
Solution: \begin{align*}x = 66\end{align*}
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.
\begin{align*}\frac{x}{3}\end{align*}
Next, we know that Casey added four pounds to one of the portions.
\begin{align*}\frac{x}{3} + 4\end{align*}
Jessica ended up with 8 pounds.
\begin{align*}\frac{x}{3} + 4 = 8\end{align*}
Now we can solve the equation. Start by subtracting four from both sides of the equation.
\begin{align*}\frac{x}{3} + 4  4 = 8  4\end{align*}
\begin{align*}\frac{x}{3} = 4\end{align*}
Next we use the inverse of division, multiplication, and multiply three times four.
\begin{align*}x = 12\end{align*}
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.
\begin{align*}\frac{y}{19} + 6 = 10\end{align*}
Solution
First, we have to subtract 6 from each side of the equation.
\begin{align*}\frac{y}{19} = 10  6\end{align*}
\begin{align*}\frac{y}{19} = 4\end{align*}
Now we can multiply 19 times 4. This will give us the value of \begin{align*}y\end{align*}
\begin{align*}(19)(4) = 76\end{align*}
\begin{align*}y = 76\end{align*}
This is our solution.
Video Review
Practice
Directions: Solve the following two step equations that have addition and division in them.

\begin{align*}\frac{x}{3} + 4 = 8\end{align*}
x3+4=8  \begin{align*}\frac{x}{5} + 8 = 10\end{align*}
 \begin{align*}\frac{a}{6} + 7 = 13\end{align*}
 \begin{align*}\frac{a}{9} + 4 = 30\end{align*}
 \begin{align*}\frac{b}{8} + 6 = 15\end{align*}
 \begin{align*}\frac{c}{12} + 9 = 18\end{align*}
 \begin{align*}\frac{x}{7} + 7 = 21\end{align*}
 \begin{align*}\frac{x}{11} + 5 = 12\end{align*}
 \begin{align*}\frac{x}{12} + 9 = 16\end{align*}
 \begin{align*}\frac{a}{14} + 6 = 8\end{align*}
 \begin{align*}\frac{x}{22} + 9 = 12\end{align*}
 \begin{align*}\frac{y}{2} + 14 = 18\end{align*}
 \begin{align*}\frac{x}{7} + 24 = 38\end{align*}
 \begin{align*}\frac{x}{8} + 15 = 30\end{align*}
 \begin{align*}\frac{x}{9} + 11 = 28\end{align*}
Algebraic Equation
An algebraic equation contains numbers, variables, operations, and an equals sign.Equation
An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.OneStep Equation
A onestep equation is an algebraic equation with one operation in it that requires one step to solve.TwoStep Equation
A twostep equation is an algebraic equation with two operations in it that requires two steps to solve.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.Image Attributions
Description
Learning Objectives
Here you'll solve equations involving the inverse properties of addition and division.