Solving equations with absolute values on both sides can seem intimidating at first, but with a clear strategy the process becomes straightforward and reliable. By systematically examining these cases, you can isolate the variable and verify each potential solution. In real terms, the key idea is to recognize that an absolute value expression represents distance from zero, which leads to two possible cases for each side of the equation. This guide walks you through the reasoning, provides a step‑by‑step method, works through several examples, highlights common pitfalls, and offers tips to build confidence when tackling any absolute‑value equation where the absolute value appears on both sides.
Understanding Absolute Value Basics
The absolute value of a real number x, denoted |x|, is defined as
[ |x| = \begin{cases} x & \text{if } x \ge 0,\ -x & \text{if } x < 0. \end{cases} ]
Geometrically, |x| measures how far x lies from zero on the number line, always yielding a non‑negative result. When an absolute value appears on both sides of an equation, such as
[ |A| = |B|, ]
the equality holds precisely when the quantities inside the bars are either equal or opposites:
[ A = B \quad \text{or} \quad A = -B. ]
This dual‑possibility foundation drives the solving technique The details matter here..
General Approach to |A| = |B|
- Isolate the absolute‑value expressions (if they are not already alone on each side).
- Set up the two cases derived from the definition:
- Case 1: A = B
- Case 2: A = –B
- Solve each resulting linear (or quadratic, etc.) equation for the variable.
- Check every candidate solution in the original equation, because extraneous roots can arise when squaring or manipulating expressions.
- Accept only those solutions that satisfy the original absolute‑value equation.
If the equation contains additional terms outside the absolute values (e.g., |A| + c = |B|), first move those terms to isolate the absolute‑value blocks before applying the case split Surprisingly effective..
Step‑by‑Step Method (Illustrated)
Consider a generic equation
[ |ax + b| = |cx + d|. ]
Follow these steps:
-
Write the two cases
- Case 1: ax + b = cx + d
- Case 2: ax + b = -(cx + d)
-
Solve Case 1
[ ax + b = cx + d ;\Longrightarrow; (a-c)x = d-b ;\Longrightarrow; x = \frac{d-b}{a-c}, ] provided (a \neq c). If (a = c), the equation reduces to (b = d); either all real numbers satisfy it (if true) or none do (if false) Simple, but easy to overlook. And it works.. -
Solve Case 2
[ ax + b = -cx - d ;\Longrightarrow; (a+c)x = -d-b ;\Longrightarrow; x = \frac{-d-b}{a+c}, ] provided (a \neq -c). If (a = -c), the equation becomes (b = -d); again, either all real numbers or none satisfy it Simple as that.. -
Verify each candidate by substituting back into (|ax+b| = |cx+d|). Discard any that fail.
Worked Examples
Example 1: Simple Linear Case
Solve (|2x - 3| = |x + 4|).
Case 1: (2x - 3 = x + 4) → (x = 7).
Case 2: (2x - 3 = -(x + 4)) → (2x - 3 = -x - 4) → (3x = -1) → (x = -\frac13).
Check:
- For (x = 7): (|2·7-3| = |11| = 11); (|7+4| = |11| = 11) ✔️
- For (x = -\frac13): (|2(-\frac13)-3| = |-\frac23-3| = |-\frac{11}{3}| = \frac{11}{3}); (|-\frac13+4| = |\frac{11}{3}| = \frac{11}{3}) ✔️
Both solutions are valid: (x = 7) or (x = -\frac13) Worth knowing..
Example 2: No Solution Scenario
Solve (|5x+1| = |5x-2|).
Case 1: (5x+1 = 5x-2) → (1 = -2) (false) → no solution from this case.
Case 2: (5x+1 = -(5x-2)) → (5x+1 = -5x+2) → (10x = 1) → (x = 0.1) Worth knowing..
Check:
(|5·0.1+1| = |0.5+1| = 1.5); (|5·0.1-2| = |0.5-2| = 1.5) ✔️
Thus the equation has a single solution (x = 0.1). Notice that the first case produced a contradiction, indicating that the expressions inside the absolute values can never be equal; only the opposite‑sign case works.
Example 3: Quadratic Inside the Absolute Value
Solve (|x^2 - 4| = |2x - 3|).
Because the left side is quadratic, we still apply the two‑case method, but each case may yield a quadratic equation And that's really what it comes down to. That's the whole idea..
Case 1: (x^2 - 4 = 2x - 3) → (x^2 - 2x -1 = 0) → (x = 1 \pm \sqrt{2}).
Case 2: (x^2 - 4 = -(2x - 3)) → (x^2 - 4 = -2x + 3) → (x^2 + 2x -7 = 0) → (x = -1 \pm \sqrt{8}) → (x = -1 \pm 2\sqrt{2}).
Now test each of the four candidates in the original equation:
- (x = 1+\sqrt{2}\approx 2.414): LHS (|(2.414)^2-4| = |5.828-4| = 1.828); RHS (|2·2.414
Example 3 (continued): Verification of All Candidates
We have four algebraic candidates from the two cases:
[ x_1 = 1+\sqrt2,\qquad x_2 = 1-\sqrt2,\qquad x_3 = -1+2\sqrt2,\qquad x_4 = -1-2\sqrt2 . ]
We substitute each into the original equation (|x^{2}-4| = |2x-3|) to confirm that none are extraneous.
| Candidate | (|x^{2}-4|) (LHS) | (|2x-3|)
