# How do you solve |x + 3| + | x - 2| = 4?

Jun 8, 2017

Given: $| x + 3 | + | x - 2 | = 4$

Subtract $| x + 3 |$ from both sides:

$| x - 2 | = 4 - | x + 3 | \text{ [1]}$

Take a minute to look at application of the definition:

|A| = {(A; A >=0),(-A; A < 0):}

|x - 2| = {(x - 2; x -2 >=0),(2-x; x -2 < 0):}
|x + 3| = {(x + 3; x +3 >=0),(-x-3; x +3 < 0):}

Simplifying the inequalities:

|x - 2| = {(x - 2; x >= 2),(2-x; x < 2):}
|x + 3| = {(x + 3; x >=-3),(-x-3; x < -3):}

Substitute the case $x \ge 2$ into equation [1]:

x - 2 = 4-(x + 3);x >=2

x - 2 = 4-x - 3;x >=2

2x = 4 - 3+ 2;x >=2

2x = 3;x >=2

x = 3/2; x >=2 larr This solution is outside the domain, therefore, it is invalid.

Substitute the case $- 3 \le x < 2$ into equation [1]:

2-x = 4-(x + 3);-3<=x<2

2-x = 4-x - 3;-3<=x<2

2 = 1;-3<=x<2 larr no solution

Substitute the case $x < - 3$ into equation [1]:

2-x = 4-(-x-3);x < -3

2-x = 4+x+3;x < -3

-2x = 4+3-2;x < -3

-2x = 5;x < -3

x = -5/2;x < -3 larr This solution is outside the domain, therefore, it is invalid.

## The overall answer is that there is no solution.

I checked this with WolframAlpha and it agreed with me.