# What is the solution set for abs(2x – 3) – 8 = –1?

Aug 26, 2015

$x = - 2 \text{ }$ or $\text{ } x = 5$

#### Explanation:

Start by isolating the modulus on one side of the equation by adding $8$ to both sides

$| 2 x - 3 | - \textcolor{red}{\cancel{\textcolor{b l a c k}{8}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{8}}} = - 1 + 8$

$| 2 x - 3 | = 7$

As you know, the absolute value of a real number is always positive irrespective of that number's sign.

This tells you that you have two cases to think about, one in which the expression that's inside the modulus is positive, and the other in which the expression inside the modulus is negative.

• $2 x - 3 > 0 \implies | 2 x - 3 | = 2 x - 3$

This will make your equation take the form

$2 x - 3 = 7$

$2 x = 10 \implies x = \frac{10}{2} = \textcolor{g r e e n}{5}$

• $2 x - 3 < 0 \implies | 2 x - 3 | = - \left(2 x - 3\right)$

This time, you have

$- \left(2 x - 3\right) = 7$

$- 2 x + 3 = 7$

$- 2 x = 4 \implies x = \frac{4}{\left(- 2\right)} = \textcolor{g r e e n}{- 2}$

So there ar eactually two possible solutions to this equation, one that makes $2 x + 3$ positive, $x = 5$, and one that makes $2 x + 3$ negative, $x = - 2$.