What if you had an algebraic equation involving addition or subtraction like \begin{align*}x - \frac{2}{5} = \frac{5}{8}\end{align*}? How could you solve it for the unknown variable *x*? After completing this Concept, you'll be able to solve equations like this one.

### Watch This

CK-12 Foundation: 0301S Solving Equations with Addition and Subtraction (H264)

### Guidance

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*} to represent the cost of the mp3 player, we can write the equation \begin{align*}x + 22 = 100\end{align*}. This tells us that the value of the player **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*}. This tells us that the value of the player is **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.

We’ll start with simple problems like the one in the last example.

**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.

#### Example A

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

**Solution**

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*}

#### Example B

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

**Solution**

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!

#### Example C

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

**Solution**

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.

Watch this video for help with the Examples above.

CK-12 Foundation: Solving Equations with Addition and Subtraction

### Vocabulary

- An equation in which each term is either a constant or the product of a constant and a single variable is a
**linear equation**. - We can add, subtract, multiply, or divide both sides of an equation by the same value and still have an
**equivalent equation**. - To solve an equation,
**isolate**the unknown variable on one side of the equation by applying one or more arithmetic operations to both sides.

### Guided Practice

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

**Solution**

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*}

### Practice

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*}