Question

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

Answer 1

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

`11/7"  "<"  "x"  "<"  "17/7`.

Explanation:

Given:

`3>2abs(x-2)+5abs(x-2)`

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

`3>7abs(x-2)`

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

`3/7>abs(x-2)`

Step 3: When the absolute value of something (`x-2`) is smaller than a specified value (`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 `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

`11/7"  "<"  "x"  "<"  "17/7`.