Solve |x+1| ≤ 7-x: Absolute Value Inequality Analysis

The expression inside $|x+1|$ can be positive or negative. When $x+1 \geq 0$, we get $|x+1| = x+1$. When $x+1 < 0$, we get $|x+1| = -(x+1)$.

The right side $7-x$ represents the maximum allowed distance from -1. Since distances can't be negative, we need $7-x \geq 0$, which means $x \leq 7$.

In case 2, when $x < -1$, we get $-x-1 \leq 7-x$. Adding $x$ to both sides gives $-1 \leq 7$, which is always true regardless of $x$ value.

Case 1 gives $x \leq 3$ when $x \geq -1$. Case 2 is always satisfied when $x < -1$. But we also need $x \leq 7$ for the inequality to make sense. The intersection gives us $x \leq 3$.

Absolutely! Try $x = 0$: $|0+1| = 1$ and $7-0 = 7$, so $1 \leq 7$ ✓. Try $x = 4$: $|4+1| = 5$ and $7-4 = 3$, so $5 \leq 3$ ✗.