### One-Step Equations and Inverse Operations

Nadia is buying a new mp3 player. Peter watches her pay for the player with a $100 bill. She receives $22.00 in change, and from only this information, Peter works out how much the player cost. How much was the player?

In algebra, we can solve problems like this using an **equation**. An **equation** is an algebraic expression that involves an **equals** sign. If we use the letter \begin{align*}x\end{align*}**plus** the value of the change received is **equal** to the $100 that Nadia paid.

Another way we could write the equation would be \begin{align*}x = 100 - 22\end{align*}**equal** to the total amount of money Nadia paid \begin{align*}(100 - 22)\end{align*}. This equation is mathematically equivalent to the first one, but it is easier to solve.

In this chapter, we will learn how to solve for the variable in a one-variable linear equation. **Linear equations** are equations in which each term is either a constant, or a constant times a single variable (raised to the first power). The term linear comes from the word line, because the graph of a linear equation is always a line.

**Solving Equations Using Addition and Subtraction**

When we work with an algebraic equation, it’s important to remember that the two sides have to stay equal for the equation to stay true. We can change the equation around however we want, but whatever we do to one side of the equation, we have to do to the other side. In the introduction above, for example, we could get from the first equation to the second equation by subtracting 22 from both sides:

\begin{align*} x + 22 &= 100\\ x + 22 - 22 &= 100 - 22\\ x &= 100 - 22\end{align*}

Similarly, we can add numbers to each side of an equation to help solve for our unknown.

1. Solve \begin{align*}x - 3 = 9\end{align*}.

To solve an equation for \begin{align*}x\end{align*}, we need to **isolate** \begin{align*}x-\end{align*}that is, we need to get it by itself on one side of the equals sign. Right now our \begin{align*}x\end{align*} has a 3 subtracted from it. To reverse this, we’ll add 3—but we must add 3 to **both sides.**

\begin{align*} x - 3 &= 9\\ x - 3 + 3 &= 9 + 3\\ x + 0 &= 9 + 3\\ x &= 12\end{align*}

*2.* Solve \begin{align*}z - 9.7 = -1.026\end{align*}

It doesn’t matter what the variable is—the solving process is the same.

\begin{align*} z - 9.7 &= -1.026\\ z - 9.7 + 9.7 &= -1.026 + 9.7\\ z &= 8.674\end{align*}

Make sure you understand the addition of decimals in this example!

3. Solve \begin{align*}x + \frac{4}{7} = \frac{9}{5}\end{align*}.

To isolate \begin{align*}x\end{align*}, we need to subtract \begin{align*}\frac{4}{7}\end{align*} from both sides.

\begin{align*} x + \frac{4}{7} &= \frac{9}{5}\\ x + \frac{4}{7} - \frac{4}{7} &= \frac{9}{5} - \frac{4}{7}\\ x &= \frac{9}{5} - \frac{4}{7}\end{align*}

Now we have to subtract fractions, which means we need to find the LCD. Since 5 and 7 are both prime, their lowest common multiple is just their product, 35.

\begin{align*}x &= \frac{9}{5} - \frac{4}{7}\\ x &= \frac{7 \cdot 9}{7 \cdot 5} - \frac{4 \cdot 5}{7 \cdot 5}\\ x &= \frac{63}{35} - \frac{20}{35}\\ x &= \frac{63 - 20}{35}\\ x &= \frac{43}{35}\end{align*}

Make sure you’re comfortable with decimals and fractions! To master algebra, you’ll need to work with them frequently.

### Example

#### Example 1

Solve \begin{align*}x +10 = 17\end{align*}.

To solve an equation for \begin{align*}x\end{align*}, we need to **isolate** \begin{align*}x-\end{align*}that is, we need to get it by itself on one side of the equals sign. Right now our \begin{align*}x\end{align*} has 10 added to it. To reverse this, we’ll subtract 10—but we must subtract 10 to **both sides.**

\begin{align*} x +10 &= 17\\ x +10- 10 &= 17 -10\\ x + 0 &= 17-10\\ x &= 7\end{align*}

### Review

For 1-5, solve the following equations for \begin{align*}x\end{align*}.

- \begin{align*}x - 11 = 7\end{align*}
- \begin{align*}x - 1.1 = 3.2\end{align*}
- \begin{align*}x +0.257 = 1\end{align*}
- \begin{align*}x + \frac{5}{2} = \frac{2}{3}\end{align*}
- \begin{align*}x - \frac{5}{6} = \frac{3}{8}\end{align*}

For 6-10, solve the following equations for the unknown variable.

- \begin{align*}q - 13 = -13\end{align*}
- \begin{align*}z + 1.1 = 3.0001\end{align*}
- \begin{align*}r + 1 = \frac{2}{5}\end{align*}
- \begin{align*}t + \frac{1}{2} = \frac{1}{3} \end{align*}
- \begin{align*}\frac{3}{4} = -\frac{1}{2} - y \end{align*}

### Review (Answers)

To view the Review answers, open this PDF file and look for section 3.1.