Solve the following two absolute value equations: g) $|x-1| + |x-2| + |x-3| = 18$ i) $2 + |x-6| = |x-4|$

1. Problem Description

Solve the following two absolute value equations:

g) $|x-1| + |x-2| + |x-3| = 18$

i) $2 + |x-6| = |x-4|$

2. Solution Steps

g) $|x-1| + |x-2| + |x-3| = 18$

We consider the critical points $x=1, x=2, x=3$.

Case 1: $x < 1$. Then $1-x + 2-x + 3-x = 18$, so $6-3x=18$, $3x = -12$, and $x=-4$. Since $-4 < 1$, $x=-4$ is a solution.

Case 2: $1 \leq x < 2$. Then $x-1 + 2-x + 3-x = 18$, so $4-x = 18$, $x = -14$. Since $-14$ is not in $[1,2)$, there is no solution in this case.

Case 3: $2 \leq x < 3$. Then $x-1 + x-2 + 3-x = 18$, so $x = 18$. Since $18$ is not in $[2,3)$, there is no solution in this case.

Case 4: $x \geq 3$. Then $x-1 + x-2 + x-3 = 18$, so $3x-6 = 18$, $3x = 24$, and $x=8$. Since $8 \geq 3$, $x=8$ is a solution.

Therefore, the solutions are $x=-4$ and $x=8$.

i) $2 + |x-6| = |x-4|$

We consider the critical points $x=4, x=6$.

Case 1: $x < 4$. Then $2 + 6-x = 4-x$, so $8-x = 4-x$, $8=4$, which is impossible. Thus, there are no solutions for $x < 4$.

Case 2: $4 \leq x < 6$. Then $2 + 6-x = x-4$, so $8-x = x-4$, $2x = 12$, $x=6$. Since $x < 6$, $x=6$ is not a solution.

Case 3: $x \geq 6$. Then $2 + x-6 = x-4$, so $x-4 = x-4$, which is always true. Thus, any $x \geq 6$ is a solution.

Therefore, the solution is $x \geq 6$.

3. Final Answer

g) $x = -4, x = 8$

i) $x \geq 6$