# Solve the equation: |3x+2|=|3-2x|Solve the equation: |3x+2|=|3-2x|

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

We have 4 cases:

case 1:

3x + 2 = 3- 2x

Group similar terms:

5x = 1

**==> x= 1/5**

** **

Case 2:

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

-3x - 2 = 3 - 2x

==> -x = 5

**==> x= -5**

** **

Case 3:

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

3x + 2 = -3 + 2x

==> **x= -5**

** **

Case 4:

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

-3x - 2 = -3 + 2x

==> -5x = -1

**==> x= 1/5**

**Then the solution is: **

**x= { 1/5, -5}**

We'll develop the modulus:|3x+2|

3x+2,for 3x + 2>=0

3x>=-2

x>=-2/3

-3x - 2, for 3x + 2< 0

x<-2/3

We'll develop the modulus:|3-2x|

3-2x, for 3-2x>=0

-2x>=-3

x=<3/2

2x - 3, for x>3/2

We'll discuss 3 cases:

Case 1)

x belongs to the interval (-infinite, -2/3)

-3x - 2 = 3 - 2x

We'll isolate x to the left side:

-3x + 2x = 3 + 2

-x = 5

**x = -5**

Since x belongs to the interval of admissible values, we'll accept it.

Case 2)

x belongs to the interval [-2/3 ; 3/2]

3x+2 = 3 - 2x

5x = 1

**x = 1/5**

Since x belongs to the interval of admissible values, we'll accept it.

Case 3)

x belongs to the interval (3/2 ; +infinite)

3x + 2 = 2x - 3

x = -5

We'll reject the solution since x doesn't belong to the established interval.

|3x+2| = |3-2x|

To find the solutions.

Solution:

|a| = positive value of a.

LHS vanishes for x = -2/3 and RHS vanishes for x = 3/2. So we consider 3 cases: (i)x <-2/3 , (ii) -2/3 < x < 3/2 and (iii) x > 3/2.

So , when x < -2/3, 3x+2 < 0 , |3x+2| has the positive value |3x+2| = -(3x+2) and |3-2x| = 2-3x.

Therefore -3x-2= 3-2x . Or -3x+2x = 3+2 , Or

** x = -5.** So when x < -2/3, the solution is x= -5

When x>-2/3 and x < 3/2, both LHS and RHS are positive.

| 3x+2| = 3x+2 and |3-2x| = 3-2x. So,

3x+2 = 3-2x

3x+2x = 3-2 = 1

5x =1

**x = 1/5, when -2/3 < x < 3/2**

When x > 3/2,

|3x+2| positve and |3-2x| is negative. So

3x+2 = 2x-3

3x-2x = -3-2 = -5

x= -5 seems valid by bysubstitution in the given equation, but stiill not valid , as x >3/2 by assumption and the solution x = -5 cannot go together.

So the solution is : When x = -5 and x = 1/5.

|3x+2| denotes the absolute value of 3x+2 and |3-2x| denotes the absolute value of 3-2x.

Now we have 4 conditions which are possible:

- 3x+2 and 3-2x are negative:

For this case, 3x+2=3-2x => 5x=1 => x=1/5

- 3x+2 and 3-2x are positive:

Here, 3x+2= 3-2x => 5x=1 => x=1/5

- 3x+2 is negative and 3-2x is positive:

Here, 3x+2= 3-2x => 5x=1 => x=1/5

- 3x+2 is positive and 3-2x is negative:

Here 3x+2=2x-3 => x=-5

**Therefore the valid values of x are -5 and 1/5**