# How do you solve  3 > 2|x-2|+5|x-2|?

Aug 8, 2017

Combine like terms; isolate the absolute value; solve for it; rewrite the inequality without the absolute value signs; isolate $x$.

$\frac{11}{7} \text{ "<" "x" "<" } \frac{17}{7}$.

#### Explanation:

Given:

$3 > 2 \left\mid x - 2 \right\mid + 5 \left\mid x - 2 \right\mid$

Step 1: Since the value inside the absolute brackets is the same for both terms on the RHS, we can add them together:

$3 > 7 \left\mid x - 2 \right\mid$

Step 2: We can now isolate the absolute value brackets by dividing both sides by 7, like this:

$\frac{3}{7} > \left\mid x - 2 \right\mid$

Step 3: When the absolute value of something ($x - 2$) is smaller than a specified value ($\frac{3}{7}$), that means its magnitude must be less than that value. In this case, that means $x - 2$ can be positive or negative, but it must be within $\frac{3}{7}$ units of 0 either way. We can rewrite this as a new inequality, thereby eliminating the absolute value brackets:

–3/7"  "<"  "x-2"  "<"  "3/7

Step 4: Isolate $x$; add 2 to all three "sides".

–3/7+2"  "<"  "x"  "<"  "3/7+2

which reduces to

$\frac{11}{7} \text{ "<" "x" "<" } \frac{17}{7}$.