Answer
No solution :
$|x - 2| + 6 = 0$
$|x - 6| - 5 = -9$
One solution:
$|x - 1| + 4 = 4$
$|x - 5| - 8 = -8$
Two solutions :
$|x + 3| - 1 = 0$
$|x + 8| + 2 = 7$
Work Step by Step
Let's consider the absolute value equation $$|x-h|=d.\tag{1}$$ $\textbf{Case 1: No solution}$
The equation $(1)$ cannot be solved if the right hand side is a negative number ($d<0$) as absolute values are always positive.
a) $|x - 2| + 6 = 0$
Subtract $6$ from both sides $$|x - 2| = -6.$$ As RHS is negative, the equation has no solution.
b) $|x - 6| - 5 = -9$
Add $5$ to both sides $$|x-6| = -4$$ As $RHS$ is negative, the equation has no solution.
$\textbf{Case 2: One solution}$
The equation $(1)$ will have one solution if the right hand side is $0$ as zero cannot assume any sign ($d=0$).
a) $|x - 1| + 4 = 4$
Subtract $4$ from both sides $$|x-1| = 0.$$ As $RHS = 0$, we have one solution.
b) $|x - 5| - 8 = -8$
Add $8$ to both sides $$|x - 5| = 0.$$ As $RHS = 0$, the equation has one solution.
$\textbf{Case 3: Two solutions}$
If the right hand side of the solution is a real number greater than $0$ ($d>0$), then it has $2$ solutions.
a) $|x + 3| - 1 = 0$
Add $1$ to both sides $$|x+3| = 1.$$ As $RHS =1 >0$, there are $2$ solutions.
b) $|x + 8| + 2 = 7$
Subtract $2$ from both sides $$|x+8| = 5.$$ As $RHS = 5 >0$, this equation has $2$ solutions.
