# Doing College Algebra and came across a Tricky Problem?

l(3x-2)divided by 5l is less then or equal to 1/2

and i'm doing inequalities in one variable (intermediate algebra eight edition)the lines on both sides are absolute value symbols

Yes it is the second thing you said

Thank you

### 3 Answers

- billrussell42Lv 71 month agoFavourite answer
is that

l(3x-2)/5l ≤ 1/2

but what is the question?

solve for x?

((3x-2)/5)² ≤ 1/4

(9x² – 12x + 4)/25 ≤ 1/4

9x² – 12x + 4 ≤ 25/4

36x² – 48x + 16 ≤ 25

36x² – 48x – 9 ≤ 0

12x² – 16x – 3 ≤ 0

(2x – 3)(6x + 1) ≤ 0

find critical points, which are x = +3/2 and –1/6

Check intervals in between critical points.

(Test values in the intervals to see if they work.)

x ≤ –1/6

l(3x-2)/5l ≤ 1/2

l(3(–1)-2)/5l ≤ 1/2

l(–5)/5l ≤ 1/2

1 ≤ 1/2 NO

(Doesn't work in original inequality)

–1/6 ≤ x ≤ 3/2

l(3x-2)/5l ≤ 1/2

l(3(1)-2)/5l ≤ 1/2

l(1)/5l ≤ 1/2

1/5 ≤ 1/2 ok

(Works in original inequality)

x ≥ 3/2

l(3•2-2)/5l ≤ 1/2

l(4)/5l ≤ 1/2

4/5 ≤ 1/2 NO

(Doesn't work in original inequality)

answer

–1/6 ≤ x ≤ 3/2

- PuzzlingLv 71 month ago
Here's how I interpreted your inequality:

|(3x - 2)/5| ≤ 1/2

If you take off the absolute value symbols, the value must be between -1/2 and 1/2:

-1/2 ≤ (3x - 2)/5 ≤ 1/2

We're trying to isolate x in the middle, so start by multiplying everything by 5:

-5/2 ≤ 3x - 2 ≤ 5/2

Add 2 everywhere:

2 - 5/2 ≤ 3x ≤ 2 + 5/2

Simplify the outside values:

4/2 - 5/2 ≤ 3x ≤ 4/2 + 5/2

-1/2 ≤ 3x ≤ 9/2

Divide everything by 3 to finally get x alone:

-1/6 ≤ x ≤ 9/6

Reduce the fraction:

-1/6 ≤ x ≤ 3/2

Or:

-1/6 ≤ x ≤ 1½

- Anonymous1 month ago
Oh I did that a few years ago but don't remember anything sorry :(