# Topic 9 Absolute Value Equations

## 9.1 The Direction of a Number

Can you determine the value of the expression \(\dfrac{|x|}{x}\) for all nonzero real number \(x\) and explain the meaning of the value?

## 9.2 Properties of Absolute Values

The ** absolute value** of a real number \(a\), denoted by \(|a|\), is the distance from \(0\) to \(a\) on the number line. In particular, \(|a|\) is always greater than or equal to \(0\), that is \(|a|\geq 0\). Absolute values satisfy the following properties:
\[
|-a|=|a|, \quad |ab|=|a||b| \quad \text{and} \quad \left|\frac{a}{b}\right|=\frac{|a|}{|b|}.
\]

An absolute value equation may be rewritten as \(|X|=c\), where \(X\) represents an algebraic expression.

If \(c\) is **positive**, then the equation \(|X|=c\) is equivalent to {\(X=c\)or \(X=-c\).}

If \(c\) is **negative**, then the solution set of \(|X|=c\) is **empty**.
An ** empty set** is denoted by \(\emptyset\).

More generally, \(|X|=|Y|\) is equivalent to \(X=Y\) or \(X=-Y\).

The equation \(|X|=0\) is equivalent to \(X=0\).

**Example 9.1 **
Solve the equation
\[|2x-3|=7.\]

*Solution. *

The equation is equivalent to \[ \begin{aligned} 2x-3 =-7 & \qquad\text{or} & 2x-3 =7 \\ 2x =-4 & & 2x =10 \\ x =-2 & \qquad\text{or} & x =5 \end{aligned} \]

The solutions are \(x=-2\) or \(x=5\). In set-builder notation, the solution set is \(\{-2, 5\}\).

**Example 9.2 **
Solve the equation
\[|2x-1|-3=8.\]

*Solution. *

Rewrite the equation into \(|X|=c\) form. \[|2x-1|=11\]

Solve the equation. \[ \begin{aligned} 2x-1 =-11 & \qquad\text{or} & 2x-1 =11 \\ 2x =-10 & & 2x =12 \\ x =-5 & \qquad\text{or} & x =6 \end{aligned} \]

The solutions are \(x=-5\) or \(x=6\). In set-builder notation, the solution set is \(\{-5, 6\}\).

**Example 9.3 **
Solve the equation
\[3|2x-5|=9.\]

*Solution. *

Rewrite the equation into \(|X|=c\) form. \[|2x-5|=3\]

Solve the equation. \[ \begin{aligned} 2x-5 =-3 & \qquad\text{or} & 2x-5 =3 \\ 2x =2 & & 2x =8 \\ x =1 & \qquad\text{or} & x =4 \end{aligned} \]

The solutions are \(x=1\) or \(x=4\). In set-builder notation, the solution set is \(\{1, 4\}\).

**Example 9.4 **
Solve the equation
\[2|1-2x|-3=7.\]

*Solution. *

Rewrite the equation into \(|X|=c\) form. \[|2x-1|=5\]

Solve the equation. \[ \begin{aligned} 2x-1 =-5 & \qquad\text{ or } & 2x-1 =5 \\ 2x =-4 & & 2x =6 \\ x =-2 & \qquad\text{or} & x =3 \end{aligned} \]

The solutions are \(x=-2\) or \(x=3\). In set-builder notation, the solution set is \(\{-2, 3\}\).

**Example 9.5 **
Solve the equation
\[|3x-2|=|x+2|.\]

*Solution. *

Note that two numbers have the same absolute value only if they are the same or opposite to each other. Then the equation is equivalent to

\[3x-2=x+2\quad \text{or} \quad 3x-2=-(x+2).\]
\[
\begin{aligned}
3x-2 =x+2 & \qquad\text{or} & 3x-2 =-(x+2) \\
2x =4 & & 4x =0 \\
x =2 & \qquad\text{or} & x =0
\end{aligned}
\]

The solutions are \(x=2\) and \(x=0\). In set-builder notation, the solution set is \(\{0, 2\}\).

**Example 9.6 **
Solve the equation
\[2|1-x|=|2x+10|.\]

*Solution. *

Since \(2\) is positive, \(2|1-x|=|2||1-x|=|2-2x|\). Moreover, because \(|-X|=|X|\), the equation is equivalent to \[|2x-2|=|2x+10|.\] \[ \begin{aligned} 2x-2 =2x+10 & \qquad\text{or} & 2x-2 =-(2x+10) \\ -2 =10 & \qquad\text{or} & 4x =-8 \\ & & x =-2 \end{aligned} \]

The original equation only has one solution \(x=-2\). In set-builder notation, the solution set is \(\{-2\}\).

## 9.3 Practice

**Problem 9.1 **
Find the solution set for the equation.

- \(|2x-1|=5\)
- \(\left|\dfrac{3x-9}{2}\right|=3\)

**Problem 9.2 **
Find the solution set for the equation.

- \(|3x-6|+4=13\)
- \(3|2x-5|=9\)

**Problem 9.3 **
Find the solution set for the equation.

- \(|5x-10|+6=6\)
- \(-3|3x-11|=5\)

**Problem 9.4 **
Find the solution set for the equation.

- \(3|5x - 2|-4 = 8\)
- \(-2|3x + 1| + 5= -3\)

**Problem 9.5 **
Find the solution set for the equation.

- \(|5x-12|=|3x-4|\)
- \(|x-1|=-5|(2-x)-1|\)

**Problem 9.6 **
Find the solution set for the equation.

- \(|2x-1|=5-x\)
- \(-2x=|x+3|\)