Solve the following equation |2x+3|+|x-2|=6x

Recall the definition of absolute value:

• |x| = x if x ≥ 0

• |x| = -x if x < 0

So you need to consider 4 different cases (2 absolute value expressions with 2 possible cases each).

(i) Suppose 2x + 3 < 0 and x - 2 < 0. The first inequality says x < -3/2 and the second says x < 2, so ultimately x < -3/2. Then

|2x + 3| + |x - 2| = 6x

-(2x + 3) - (x - 2) = 6x

-2x - 3 - x + 2 = 6x

-3x - 1 = 6x

9x = -1

x = -1/9

But -1/9 is not smaller than -3/2, so this case provides no valid solution.

(ii) Suppose 2x + 3 ≥ 0 and x - 2 < 0. Then x ≥ -3/2 and x < 2, or -3/2 ≤ x < 2. Under this condition,

|2x + 3| + |x - 2| = 6x

(2x + 3) - (x - 2) = 6x

2x + 3 - x + 2 = 6x

x + 5 = 6x

5x = 5

x = 1

This solution is valid because it does fall in the interval -3/2 ≤ x < 2.

(iii) Suppose 2x + 3 < 0 and x - 2 ≥ 0. Then x < -3/2 or x ≥ 2. So

|2x + 3| + |x - 2| = 6x

-(2x + 3) + (x - 2) = 6x

-2x - 3 + x - 2 = 6x

-x - 5 = 6x

7x = -5

x = -5/7

This isn't a valid solution, because neither -5/7 < -3/2 nor -5/7 ≥ 2 are true.

(iv) Suppose 2x + 3 ≥ 0 and x - 2 ≥ 0. Then x ≥ -3/2 and x ≥ 2, or simply x ≥ 2.

|2x + 3| + |x - 2| = 6x

(2x + 3) + (x - 2) = 6x

2x + 3 + x - 2 = 6x

3x + 1 = 6x

3x = 1

x = 1/3

This is yet another invalid solution since 1/3 is smaller than 2.

So there is one solution at x = 1.