Absolute value equations

The absolute value of a quantity, written $|A|$, is simply how far it is from zero on the number line. Distance is never negative, so $|A| \ge 0$ for every $A$. The most useful consequence is that an absolute value can never equal a negative number: if you ever reach a statement like $|\text{something}| = \text{negative}$, there are no real solutions.

Because the inside of the bars can be positive or negative while giving the same magnitude, every absolute-value equation splits into cases. If $|A| = B$ where $B > 0$, then either $A = B$ or $A = -B$. You solve both equations and keep all the answers.

The same idea handles two absolute values. If $|A| = |B|$, the quantities have equal magnitude, so they are either equal or opposite: $A = B$ or $A = -B$. The four sign combinations collapse into just these two distinct equations.

So the method is always the same: isolate the absolute value, check that it equals a non-negative number, then split into the two cases and solve each. This is the engine behind solving $|3x - 6| = 9$, comparing $|2x + 3| = |x - 6|$, and reasoning about expressions where the sign of the inside matters.

Solve $|2x - 4| = 10$.
Since $10 > 0$, split into two cases. From $2x - 4 = 10$ we get $2x = 14$, so $x = 7$.
From $2x - 4 = -10$ we get $2x = -6$, so $x = -3$. The solutions are $x = 7$ and $x = -3$.

Solve $|x + 1| = |2x - 5|$.
Equal absolute values mean the insides are equal or opposite. Case one: $x + 1 = 2x - 5$ gives $x = 6$.
Case two: $x + 1 = -(2x - 5) = -2x + 5$ gives $3x = 4$, so $x = \frac{4}{3}$. The solutions are $x = 6$ and $x = \frac{4}{3}$.

Solve $|x + 2| = -5$.
An absolute value is always $\ge 0$, so it can never equal $-5$.
There is no solution, and no case-splitting is needed.