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}
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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
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