In short

When both sides of an equation are absolute values — |f(x)| = |g(x)| — you only get two cases, not four:

  1. f(x) = g(x)
  2. f(x) = -g(x)

The other two sign combinations you might be tempted to write (-f = g and -f = -g) are not new equations. Multiplying both sides by -1 doesn't change the solution set, so -f = g is the same as f = -g, and -f = -g is the same as f = g. Two equations, solve each, done. You can also see this with the squaring trick: |f|^2 = |g|^2 \Rightarrow f^2 = g^2 \Rightarrow (f-g)(f+g) = 0 — which gives exactly those two cases.

You are sitting with |x - 1| = |2x + 3| in front of you. Your brain immediately wants to do the safe thing: pop the bars off both sides, but be careful — each side could be positive or negative. So you start writing.

+(x-1) = +(2x+3)
+(x-1) = -(2x+3)
-(x-1) = +(2x+3)
-(x-1) = -(2x+3)

Four cases. Four equations to solve. That's a lot of arithmetic for one homework problem.

Here's the good news: half of those equations are duplicates. You only ever need to solve two of them. Once you see why, this whole class of problem (CBSE Class 11 loves it) becomes a 30-second job.

The two-case rule

When you see the equation |f(x)| = |g(x)|, think about what it is really saying. Both sides have the bars stripped to magnitude. So the equation is claiming:

"The distance of f(x) from zero equals the distance of g(x) from zero."

Two real numbers have the same distance from zero in exactly two ways: they are the same number, or they are exact negatives of each other (mirror images across zero). 7 and 7 both sit 7 units from zero. 7 and -7 also both sit 7 units from zero. There is no third way.

So the equation |f(x)| = |g(x)| unfolds into:

\boxed{\,f(x) = g(x) \quad \text{or} \quad f(x) = -g(x)\,}

That is the entire rule. Two cases. Solve both linear equations, collect the roots, you are done.

Unfolding |f(x)| = |g(x)| into two cases, not fourA diagram showing the original equation at the top splitting into two boxes below: f equals g on the left and f equals minus g on the right. Two greyed-out boxes labelled minus f equals g and minus f equals minus g are shown crossed out, with arrows indicating each is equivalent to one of the kept cases. |f(x)| = |g(x)| f(x) = g(x) f(x) = −g(x) −f(x) = −g(x) −f(x) = g(x) same as f = g same as f = −g Four sign combinations collapse into two distinct equations.
The four sign possibilities you might be tempted to write are not four equations — they are two equations written twice.

Why not four cases?

Look at one of the discarded equations: -f(x) = g(x). Multiply both sides by -1:

-f(x) = g(x) \quad \xRightarrow{\times (-1)} \quad f(x) = -g(x)

That's case 2 from above. Why: multiplying both sides of an equation by the same non-zero number doesn't change which x values make it true — it's the same equation, just rearranged. So -f = g and f = -g have identical solution sets.

Now the other one: -f(x) = -g(x). Multiply both sides by -1 again:

-f(x) = -g(x) \quad \xRightarrow{\times (-1)} \quad f(x) = g(x)

That's case 1. Why: flipping the sign on both sides simultaneously cancels out — the two negatives mirror each other, leaving the original positive equation untouched.

So the four-way table you wrote at the start collapses cleanly:

Sign on left Sign on right Equation Same as
+ + f = g case 1
+ - f = -g case 2
- + -f = g case 2 (×−1)
- - -f = -g case 1 (×−1)

Two genuinely different equations. The other two are just relabelled twins.

The square-both-sides shortcut

If you don't trust the case argument, here's an algebraic proof that uses no case-splitting at all. Recall from the parent article that |x|^2 = x^2 for every real x (since squaring already kills the sign). So:

|f(x)| = |g(x)| \quad \Longleftrightarrow \quad |f(x)|^2 = |g(x)|^2 \quad \Longleftrightarrow \quad f(x)^2 = g(x)^2

Why this is reversible: both sides are non-negative, so squaring doesn't introduce phantom solutions the way it does in \sqrt{x} = -3 type equations. Every solution of the squared form is a solution of the original.

Now move everything to one side and factor as a difference of squares:

f(x)^2 - g(x)^2 = 0
\bigl(f(x) - g(x)\bigr)\bigl(f(x) + g(x)\bigr) = 0

A product is zero exactly when one of the factors is zero. So either f(x) - g(x) = 0, giving f = g, or f(x) + g(x) = 0, giving f = -g. Same two cases, derived without ever opening the absolute value bars. This is sometimes the cleanest method on a CBSE board paper because there's no case bookkeeping — just one factorisation.

Worked examples

The original puzzle: $|x - 1| = |2x + 3|$

Case 1: x - 1 = 2x + 3 Subtract x from both sides: -1 = x + 3. Subtract 3: x = -4.

Case 2: x - 1 = -(2x + 3) Distribute the minus: x - 1 = -2x - 3. Add 2x: 3x - 1 = -3. Add 1: 3x = -2, so x = -\tfrac{2}{3}.

Solutions: x = -4 and x = -\tfrac{2}{3}.

Quick check at x = -4: |{-4} - 1| = |{-5}| = 5 and |2(-4) + 3| = |{-5}| = 5. Match. Quick check at x = -\tfrac{2}{3}: |{-\tfrac{2}{3}} - 1| = |{-\tfrac{5}{3}}| = \tfrac{5}{3} and |2(-\tfrac{2}{3}) + 3| = |\tfrac{5}{3}| = \tfrac{5}{3}. Match.

A second one: $|3x - 5| = |x + 1|$

Case 1: 3x - 5 = x + 1 Subtract x: 2x - 5 = 1. Add 5: 2x = 6, so x = 3.

Case 2: 3x - 5 = -(x + 1) Distribute: 3x - 5 = -x - 1. Add x: 4x - 5 = -1. Add 5: 4x = 4, so x = 1.

Solutions: x = 3 and x = 1.

Verify by plugging back (always do this)

For the second example, check x = 3 in the original: |3(3) - 5| = |4| = 4 and |3 + 1| = |4| = 4. Tick.

Check x = 1: |3(1) - 5| = |{-2}| = 2 and |1 + 1| = |2| = 2. Tick.

Why bother verifying when both cases are equivalent to the original? Because algebra slips happen — a sign flip in case 2 is the most common mistake on this kind of question. Plugging back catches it instantly. On a CBSE Class 11 paper you get the marks for the verification line too.

What to remember

  • |f(x)| = |g(x)| is always two cases: f = g and f = -g.
  • The "missing" two cases are not missing — they are the same equations with both sides multiplied by -1.
  • If case-splitting feels error-prone, square both sides and factor: f^2 - g^2 = (f-g)(f+g) = 0 gives the same two solutions.
  • Always plug your answers back into the original equation. It catches sign mistakes and costs you one line of work.

This rule applies anywhere you have bars equalling bars — even if f and g are quadratic or messier. The number of cases is determined by the number of separate absolute-value blocks on each side, not by their internal complexity. Two blocks, one on each side, two cases.

For more general absolute-value behaviour (one-sided equations, no-solution cases, the distance interpretation) head back to Absolute Value — Equations. For what happens when you replace the equals sign with an inequality, see Absolute Value — Inequalities.

References

  1. NCERT Class 11 Mathematics, Chapter 6 — Linear Inequalities and the Modulus Function. NCERT textbook portal.
  2. Stewart, Redlin, Watson, Precalculus: Mathematics for Calculus, §1.8 (Absolute Value Equations and Inequalities). Publisher page.
  3. Paul's Online Math Notes — Absolute Value Equations.
  4. Khan Academy — Equations with two absolute values.
  5. CBSE sample papers, Class 11 Mathematics — past five-year question banks. CBSE academic portal.