The problem is to solve the following absolute value equations: a) $|x-2| + |x+5| = 7$ b) $|x+4| - |x-7| = 11$ c) $|x-1| - |x+5| = 6$ d) $|x+4| - |x-4| = 8$ e) $||x+2| - 5| = 8$ f) $||x-6| - 8| = 10$

1. Problem Description

The problem is to solve the following absolute value equations:

a) $|x-2| + |x+5| = 7$

b) $|x+4| - |x-7| = 11$

c) $|x-1| - |x+5| = 6$

d) $|x+4| - |x-4| = 8$

e) $||x+2| - 5| = 8$

f) $||x-6| - 8| = 10$

2. Solution Steps

a) $|x-2| + |x+5| = 7$

We consider three cases:

Case 1: $x < -5$. Then $-(x-2) - (x+5) = 7$, so $-x+2-x-5=7$, $-2x-3=7$, $-2x=10$, $x=-5$. However, we assumed $x < -5$, so $x=-5$ is not a solution.

Case 2: $-5 \le x \le 2$. Then $-(x-2) + (x+5) = 7$, so $-x+2+x+5=7$, $7=7$. Thus, all $x$ in the interval $[-5, 2]$ are solutions.

Case 3: $x > 2$. Then $(x-2) + (x+5) = 7$, so $2x+3=7$, $2x=4$, $x=2$. However, we assumed $x > 2$, so $x=2$ is not a solution.

b) $|x+4| - |x-7| = 11$

We consider three cases:

Case 1: $x < -4$. Then $-(x+4) - (-(x-7)) = 11$, so $-x-4+x-7=11$, $-11=11$, which is false. No solutions in this case.

Case 2: $-4 \le x \le 7$. Then $(x+4) - (-(x-7)) = 11$, so $x+4+x-7=11$, $2x-3=11$, $2x=14$, $x=7$.

Case 3: $x > 7$. Then $(x+4) - (x-7) = 11$, so $x+4-x+7=11$, $11=11$. Thus, all $x$ in the interval $(7, \infty)$ are solutions.

Combining cases, $x \ge 7$.

c) $|x-1| - |x+5| = 6$

Case 1: $x < -5$. Then $-(x-1) - (-(x+5)) = 6$, so $-x+1+x+5=6$, $6=6$. Thus all $x$ in the interval $(-\infty, -5)$ are solutions.

Case 2: $-5 \le x \le 1$. Then $-(x-1) - (x+5) = 6$, so $-x+1-x-5=6$, $-2x-4=6$, $-2x=10$, $x=-5$.

Case 3: $x > 1$. Then $(x-1) - (x+5) = 6$, so $x-1-x-5=6$, $-6=6$, which is false. No solutions in this case.

Combining cases, $x \le -5$.

d) $|x+4| - |x-4| = 8$

Case 1: $x < -4$. Then $-(x+4) - (-(x-4)) = 8$, so $-x-4+x-4=8$, $-8=8$, which is false. No solutions.

Case 2: $-4 \le x \le 4$. Then $(x+4) - (-(x-4)) = 8$, so $x+4+x-4=8$, $2x=8$, $x=4$.

Case 3: $x > 4$. Then $(x+4) - (x-4) = 8$, so $x+4-x+4=8$, $8=8$. Thus all $x$ in the interval $(4, \infty)$ are solutions.

Combining cases, $x \ge 4$.

e) $||x+2| - 5| = 8$

This means $|x+2| - 5 = 8$ or $|x+2| - 5 = -8$.

Case 1: $|x+2| - 5 = 8$. Then $|x+2| = 13$. So $x+2=13$ or $x+2=-13$. Thus $x=11$ or $x=-15$.

Case 2: $|x+2| - 5 = -8$. Then $|x+2| = -3$. No solution, since the absolute value must be non-negative.

Thus, $x=11$ or $x=-15$.

f) $||x-6| - 8| = 10$

This means $|x-6| - 8 = 10$ or $|x-6| - 8 = -10$.

Case 1: $|x-6| - 8 = 10$. Then $|x-6| = 18$. So $x-6=18$ or $x-6=-18$. Thus $x=24$ or $x=-12$.

Case 2: $|x-6| - 8 = -10$. Then $|x-6| = -2$. No solution, since the absolute value must be non-negative.

Thus, $x=24$ or $x=-12$.

3. Final Answer

a) $-5 \le x \le 2$

b) $x \ge 7$

c) $x \le -5$

d) $x \ge 4$

e) $x = 11$ or $x = -15$

f) $x = 24$ or $x = -12$