Solving inequalities with absolut values

Question

I have this problem:

Find the numbers x ∈ ℝ such that:

|x + 1| + |x - 4| > 7

I have already checked when both of the absolute values change its sign, but I don't know how to continue. My last step was:

x + 1 if x ≥ -1 // -x -1 if x < -1 // x-4 if x ≥ 4 // -x -4 if x<4

Answer 1

Suppose $x\ge 4$. This means neither $x+1$ nor $x-4$ are negative, so the inequality becomes $$x+1+x-4> 7$$, which simplifies to $2x\ge10$, and $x\ge 5$. So the inequality holds when $x> 5$.

Suppose $-1\le x<4$, so $x+1\ge 0$, but $x-4<0$. The inequality becomes $$x+1-x+4> 7$$ which simplifies to $5> 7$ and that's obviously not true.

Suppose $x<-1$. This means both $x+1$ and $x-4$ are negative, so the inequality becomes: $$-x-1-x+4> 7$$ Which simplifies to $-2x> 4$, so $x< -2$.

We conclude the inequality holds for all $x> 5$ and all $x< -2$

Answer 2

1. $x < -1$. $-x - 1 + 4 - x > 7 \implies 3 - 2x > 7 \implies x < -2$.
2. $x = -1$ not feasible
3. $x > -1, x < 4$. $x + 1 + 4 - x > 7$ not feasible.
4. $x = 4$ not feasible
5. $x > 4$. $x + 1 + x - 4 > 7 \implies 2x - 3 > 7 \implies x > 5$.

Thus we have $x \in (-\infty, -2) \cup (5, \infty)$.

Answer 3

You are correct to look at this with cases. Organizing the cases a bit better we have that:

$$|x+1|+|x-4|=\begin{cases}x+1+x-4&\text{if}~x\geq 4\\x+1-(x-4)&\text{if}~-1\leq x<4\\ -(x+1)-(x-4)&\text{if}~x<-1\end{cases}$$

Simplifying:

$$|x+1|+|x-4|=\begin{cases}2x-3&\text{if}~x\geq 4\\5&\text{if}~-1\leq x<4\\ -2x+3&\text{if}~x<-1\end{cases}$$

Now, look at each of these cases individually and when they are or aren't greater than $7$.