# How to solve inequalities with absolute value?I actually know how to solve it for the one problem I have has fraction in it so can someone help me solve it step by step? Problem: lx-3 over 2l +2...

How to solve inequalities with absolute value?

I actually know how to solve it for the one problem I have has fraction in it so can someone help me solve it step by step?

Problem: lx-3 over 2l +2 <6

Thanks

Asked on by boscow

### 3 Answers |Add Yours

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

l (x-3)/2l + 2 < 6

= l (x-3)/2l < 4

Then we have 2 cases:

First:

-(x-3)/2 < 4

Multiply by 2:

==> (x-3) < 8

Now add 3 to both sides:

==> x < 11

Then x belongs to (-inf, 11)

Secoond case:

-(x-3)/2 < 4

Multiply by 2:

==> -(x-3) < 8

Now multiply by (-1) , note that when multiplying by a negative number , we need to reverse the inequality:

==> (x-3) > -8

Now add 3 to both sides:

==> x > -5

Then x belongs to (-5, inf)

Then from both cases, we conclude that :

x belongs to (-5, inf) and ( -inf, 11) = (-5, 11)

The solution is x belongs to (-5, 11)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-write lx-3 over 2l +2 <6 as l(x-3) / 2l +2 <6.

Now, we'll subtract 2 both sides:

l(x-3) / 2l +2-2 < 6-2

l(x-3) / 2l < 4

This inequality implies the followings:

-4 < (x-3) / 2 < 4 <=>  l(x-3) / 2l < 4

We'll solve the left side of the inequlity:

-4 < (x-3) / 2

We'll multiply, both sides, by 2:

-4*2 < (x-3)

-8 < x - 3

We'll add 3 both sides:

-8 + 3 < x - 3 + 3

-5 < x

Now, we'll solve the right side of the inequality:

(x-3) / 2 < 4

We'll multiply by 2:

x - 3 < 8

We'll add 3 both sides:

x < 8+3

x < 11

So, the interval of admissible values for x is (-5 , 11).

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve  |x-3 over|+2 <6

Solution:

|(x-3)/2| +2 < 6.

Subtract 2 from both sides:

|(x-3)/2| +2 - 2 < 6-2

|(x-3)/2| <  4.

Case i:

If (x-3) > 3, then  (x-3)/2 < 4. Multiply by 2 both sides:

x-3 < 4*2 = 8 . Add 3 to both sides:

x < 8+3 =11

x <11............................................................(1)

Case ii:

If x -3 < 0, then LHS |x-3)/2| = (3-x)/2. So the inequality becomes:

(3-x)/2 +2 < 6. Subtract 2 from both sides:

(3-x)/2 <  6-2

(3-x)/2 <  4. Multiply by 4 both sides:

3-x < 4*2 = 8

3-x  < 8.  Subtract 8 from both sides:

3-8 -x < 8-8.

-11-x < 0. Add x to both sides:

-11< x. Or

x > -3..........................................(2).

Combining the inequalities at (1) and (2), we get:

-11 < x < 11

So x  is a number  between -11 and 11.

Or x belongs to the open interval (-11 , 11)  or x belongs to  the open interval  ] -11 ,11[ .

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