|x - 3| + |2x + 8| > 7
x -3 changes sign at x = 3 and 2x + 8 changes sign at x = -4
Thus, we consider x >= 3, -4 < x < 3 and x <= -4
When x <= -4, we have:
-(x - 3) + -(2x + 8) > 7
-x + 3 - 2x - 8 > 7
-3x - 5 > 7
3x < -12
x < -4
Thus, x < -4 is a solution.
Suppose -4 < x < 3. We have:
-(x - 3) + (2x + 8) > 7
-x + 3 + 2x + 8 > 7
x + 11 > 7
x > -4
Thus, -4 < x < 3 is a solution.
Suppose x >= 3. We have:
(x - 3) + (2x + 8) > 7
x - 3 + 2x + 8 > 7
3x + 5 > 7
3x > 2
x > 2/3
Thus, x >= 3 is a solution.
Combining all three inequalities together, we can see that the solution is actually all real numbers, except x = -4.
Solution := { x ∈ R | x =/= -4}