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

hala718 | Certified Educator

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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}

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giorgiana1976 | Student

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.

neela | Student

|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.

william1941 | Student

|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