Lesson 4: Solving Absolute Value Equations

Property

For any algebraic expression, $u$, and any positive real number, $a$, if $|u| = a$, then $u = -a$ or $u = a$.
To solve an absolute value equation, first isolate the absolute value expression.
Then, write two equivalent equations and solve each one separately.

Examples

• To solve $|3x - 1| - 5 = 10$, first isolate the absolute value: $|3x - 1| = 15$. Then set up two equations: $3x - 1 = 15$ or $3x - 1 = -15$. Solving gives $x = \frac{16}{3}$ or $x = -\frac{14}{3}$.
• The equation $3|x - 8| + 11 = 5$ simplifies to $3|x - 8| = -6$, or $|x - 8| = -2$. Since an absolute value cannot be negative, there is no solution.
• To solve $|x + 4| = |2x - 2|$, set up two cases: $x + 4 = 2x - 2$ or $x + 4 = -(2x - 2)$. Solving these yields $x = 6$ or $x = -\frac{2}{3}$.

Explanation

If the absolute value of something is a positive number $a$, it means the 'something' inside is either $a$ units to the right of zero or $a$ units to the left. This is why you must solve two separate cases.

Property

When solving an absolute value equation of the form $|ax + b| = c$ where $c < 0$, the equation has no solution because absolute values are always non-negative: $|ax + b| \geq 0$ for all real values.

Examples

Property

The equation $|ax + b| = c$ (where $c > 0$) is equivalent to:

Examples

• To solve $|x - 3| = 8$, we set up two equations: $x - 3 = 8$ or $x - 3 = -8$. The solutions are $x = 11$ and $x = -5$.
• To solve $|2y + 5| = 11$, we set up two equations: $2y + 5 = 11$ or $2y + 5 = -11$. The solutions are $y = 3$ and $y = -8$.
• To solve $|\frac{z}{3} - 1| = 4$, we set up two equations: $\frac{z}{3} - 1 = 4$ or $\frac{z}{3} - 1 = -4$. The solutions are $z = 15$ and $z = -9$.

Explanation

To solve an absolute value equation, you split it into two separate linear equations. This is because the expression inside the absolute value bars could be either positive or negative, and both would result in the same positive value.