l 3x + 1 l = l 1-2x l

we have 4 cases:

case 1:

3x + 1 = 1- 2x

==> 5x = 0

==> **x1= 0**

case 2:

3x+1 = -(1-2x)

3x +1 = -1 + 2x

==> **x2 = -2**

** **

case 3:

-(3x+1) = 1-2x

-3x -1 = 1-2x

-x = 2

==> **x3= -2**

** **

case 4:

-(3x+1) = -(1-2x)

-3x -1 = -1 + 2x

==> -x = 0

==> **x4= 0**

**Then x= { 0, -2}**

|3x+1| = |1-2x|. To solve for x.

Solution:

When 3x+1 > 0, 3x > -1 , x > -1, And 1-2x > 0 Or x <= 1/2

LHS : |3x+1 |= 3x+1, RHS : 1-2x

3x+1 = 1-2x =

5x = 0

x = 0.

When x >1/2, |3x+1| = 3x+1 and |1-2x| = 2x-1

So 3x+1 = 2x-1

3x-2x = -2

x = -2, Not consistent with the assumption.

When x < -1/3.

|3x+1| = -1-3x and |1-2x| = 1-2x.

-1-3x = 1-2x. Or

-2 = 3x-2x = x

x= -2.

So x= 0 or x = -2 are valid solutions.

We'll consider 4 cases of study.

The expressions 3x+1 and 1-2x can be either positive or negative.

Case 1:

3x+1 = 1-2x

Case 2:

-(3x+1) = 1-2x

Case 3:

3x+1= -(1-2x)

Case 4:

-(3x+1) = -(1-2x)

We notice that Case 2 and Case 3 will have the same solution Also Case 1 and Case 4 will have the same solution. So, we'll consider just Case 1 and Case 2:

Case 1:

3x+1 = 1-2x

We'll isolate x to the left side and we'll eliminate like terms:

3x + 2x = 1-1

5x = 0

We'll divide by 5:

**x = 0**

Case 2:

-(3x+1) = 1-2x

We'll remove the brackets:

-3x - 1 = 1 - 2x

We'll isolate x to the left side:

-3x + 2x = 1 + 1

-x = 2

We'll multiply by -1:

**x = -2**

**The solutions of the equation are: {-2 ; 0}.**