# How do you solve -3abs(6-x)<-15?

Aug 14, 2015

$x \in \left(- \infty , 1\right) \cup \left(11 , + \infty\right)$

#### Explanation:

First, isolate the modulus on one side of the inequality. Notice that you can do that by dividing both sides of the inequality by $- 3$ - do not forget that the sign of the inequality changes when you miltiply or divide by a negative number.

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 3}}} \cdot | 6 - x |}{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 3}}}} \textcolor{b l u e}{>} \frac{- 15}{- 3}$

$| 6 - x | > 5$

Next, use the fact that the expression inside the modulus can be positive or negative to find the solution intervals that would satisfy these possibilities.

• $6 - x > 0 \implies | 6 - x | = 6 - x$

When the expression inside the modulus is positive, you get

$6 - x > 5$

$- x > - 1 \implies x < 1$

• $6 - x < 0 \implies | 6 - x | = - \left(6 - x\right)$

When the expression inside the modulus is negative, you have

$- \left(6 -\right) > 5$

$- 6 + x > 5$

$x > 11$

So, the absolute value inequality will be true for any value of $x$ that is smaller than $1$ or for any value of $x$ that is bigger than $11$.

This means that the solution set for this inequality will be $x \in \left(- \infty , 1\right) \cup \left(11 , + \infty\right)$.