### Absolute Value Equations

In the previous concept, we saw how to solve simple absolute value equations. In this concept, you will see how to solve more complicated absolute value equations.

#### Solving Absolute Value Equations

1. Solve the equation \begin{align*}|x-4|=5\end{align*}

We consider two possibilities: the expression inside the absolute value sign is non-negative or is negative. Then we solve each equation separately.

\begin{align*}& x-4 = 5 \quad \text{and} \quad x-4=-5\\
& \quad \ \ x=9 \qquad \qquad \quad \ x=-1\end{align*}

\begin{align*}x = 9\end{align*}

The equation \begin{align*}|x-4|=5\end{align*}

2. Solve the equation \begin{align*}|x+3|=2\end{align*}

Solve the two equations:

\begin{align*}& x+3 = 2 \quad \text{and} \quad \ \ x+3=-2\\
& \quad \ \ x=-1 \qquad \qquad \quad \ x=-5\end{align*}

\begin{align*}x = -5\end{align*}

The equation \begin{align*}|x+3|=2\end{align*}

**Solve Real-World Problems Using Absolute Value Equations**

#### Real-World Application: Packing Coffee

A company packs coffee beans in airtight bags. Each bag should weigh 16 ounces, but it is hard to fill each bag to the exact weight. After being filled, each bag is weighed; if it is more than 0.25 ounces overweight or underweight, it is emptied and repacked. What are the lightest and heaviest acceptable bags?

The weight of each bag is allowed to be 0.25 ounces away from 16 ounces; in other words, the *difference* between the bag’s weight and 16 ounces is allowed to be 0.25 ounces. So if \begin{align*}x\end{align*}

Now we must consider the positive and negative options and solve each equation separately:

\begin{align*}& x-16 = 0.25 \qquad \text{and} \quad x-16 =-0.25\\
& \qquad x=16.25 \qquad \qquad \qquad \ \ x=15.75\end{align*}

**The lightest acceptable bag weighs 15.75 ounces and the heaviest weighs 16.25 ounces.**

We see that \begin{align*}16.25 - 16 = 0.25 \ ounces\end{align*}

**The answer checks out.**

The answer you just found describes the lightest and heaviest acceptable bags of coffee beans. But how do we describe the total possible range of acceptable weights? That’s where inequalities become useful once again.

### Example

#### Example 1

Solve the equation \begin{align*}|2x-7|=6\end{align*}

Solve the two equations:

\begin{align*}& 2x-7 = 6 \qquad \qquad \quad 2x-7=-6\\
& \quad \ \ 2x=13 \qquad \text{and} \qquad \ \ 2x=1\\
& \quad \ \ \ x=\frac{13}{2} \qquad \qquad \qquad \ \ x=\frac{1}{2}\end{align*}

**Answer:** \begin{align*}x=\frac{13}{2}\end{align*}

The interpretation of this problem is clearer if the equation \begin{align*}|2x-7|=6\end{align*}

### Review

Solve the absolute value equations and interpret the results by graphing the solutions on the number line.

- \begin{align*}|x-5|=10\end{align*}
|x−5|=10 - \begin{align*}|x+2|=6\end{align*}
|x+2|=6 - \begin{align*}|5x-2|=3\end{align*}
|5x−2|=3 - \begin{align*}|x-4|=-3\end{align*}
|x−4|=−3 - \begin{align*}\left|2x-\frac{1}{2}\right|=10\end{align*}
∣∣∣2x−12∣∣∣=10 - \begin{align*}|-x+5|=\frac{1}{5}\end{align*}
|−x+5|=15 - \begin{align*}\left|\frac{1}{2}x-5\right|=100\end{align*}
∣∣∣12x−5∣∣∣=100 - \begin{align*}|10x-5|=15\end{align*}
|10x−5|=15 - \begin{align*}|0.1x+3|=0.015\end{align*}
|0.1x+3|=0.015 - \begin{align*}|27-2x|=3x+2\end{align*}
|27−2x|=3x+2

### Review (Answers)

To view the Review answers, open this PDF file and look for section 6.8.