We are given two absolute value equations to solve: c) $|x-1| - |x+5| = 6$ f) $||x-6|-8| = 10$

1. Problem Description

We are given two absolute value equations to solve:

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

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

2. Solution Steps

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

We consider the critical points $x=1$ and $x=-5$. This divides the real number line into three intervals: $x < -5$, $-5 \le x < 1$, and $x \ge 1$.

Case 1: $x < -5$. Then $x-1 < 0$ and $x+5 < 0$. Thus $|x-1| = -(x-1) = 1-x$ and $|x+5| = -(x+5) = -x-5$.

So $1-x - (-x-5) = 6$, which gives $1-x+x+5 = 6$, so $6=6$. Thus any $x < -5$ is a solution.

Case 2: $-5 \le x < 1$. Then $x-1 < 0$ and $x+5 \ge 0$. Thus $|x-1| = -(x-1) = 1-x$ and $|x+5| = x+5$.

So $1-x - (x+5) = 6$, which gives $1-x-x-5 = 6$, so $-2x-4 = 6$, so $-2x = 10$, and $x = -5$.

Case 3: $x \ge 1$. Then $x-1 \ge 0$ and $x+5 > 0$. Thus $|x-1| = x-1$ and $|x+5| = x+5$.

So $x-1 - (x+5) = 6$, which gives $x-1-x-5 = 6$, so $-6 = 6$, which is impossible.

Combining the cases, we have $x \le -5$.

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

We have two cases to consider:

Case 1: $|x-6| - 8 = 10$. Then $|x-6| = 18$.

Thus $x-6 = 18$ or $x-6 = -18$.

$x = 18+6 = 24$ or $x = -18+6 = -12$.

Case 2: $|x-6| - 8 = -10$. Then $|x-6| = -2$. This is impossible since the absolute value must be non-negative.

3. Final Answer

c) $x \le -5$

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