In short

Whenever you see an equation of the shape |\text{expr}| = k, it always splits into two separate equations: \text{expr} = k and \text{expr} = -k. Solve each branch on its own. The answers you collect are the candidate solutions of the original. The number of real answers depends on k:

  • k > 0: two branches, usually two solutions.
  • k = 0: the two branches collapse into one — solve \text{expr} = 0 alone.
  • k < 0: no branches at all — distance is never negative, so no solution exists.

This unfolding move is the standard CBSE Class 11 algebraic technique for absolute value equations.

The companion article showed you the geometric picture — centre point, two arrows, two solutions. This article shows you the algebra side of the same idea: how the bars in |\text{expr}| = k literally unfold into two ordinary equations that you already know how to solve.

Think of it like this. The bars are a folded sheet of paper. As long as the paper is folded, you cannot see what is inside — the expression could be sitting at +k or at -k, and the bars hide which. The moment you unfold the paper, the two possibilities appear side by side, and each one is a clean, ordinary equation. Why: the absolute value |y| equals k exactly when y = k or y = -k. Those are the only two real numbers whose absolute value is k. So if |\text{expr}| = k, then \text{expr} has to be one of those two numbers — there is no third option to consider.

The unfolding move

Here is the rule, in one line.

|\text{expr}| = k \quad \text{(with } k > 0\text{)} \quad \Longleftrightarrow \quad \text{expr} = k \;\;\text{or}\;\; \text{expr} = -k

The double arrow matters. It says these statements are equivalent: any solution of the left equation is a solution of one of the right equations, and any solution of the right equations is a solution of the left. Nothing is lost, nothing is added.

An absolute value equation unfolding into two branches At the top, the equation absolute value of x minus 3 equals 5 sits in a rounded box. Two arrows fan downward from this box. The left arrow leads to a box containing x minus 3 equals 5. The right arrow leads to a box containing x minus 3 equals negative 5. Below each branch, a smaller box shows the solution of that branch: x equals 8 on the left, x equals negative 2 on the right. |x − 3| = 5 unfold x − 3 = 5 x − 3 = −5 solve solve x = 8 x = −2 two branches → two candidate solutions
The unfolding move on a single example. The folded equation $|x - 3| = 5$ at the top splits into two ordinary linear equations — $x - 3 = 5$ and $x - 3 = -5$ — and each is solved separately. The two answers, $x = 8$ and $x = -2$, are the solutions of the original.

The whole machinery is in that picture. The bars come off, an "or" appears, and you are left with two equations that you have known how to solve since Class 8.

Try it: unfold an equation with one click

Below is an interactive widget. The equation |x - 3| = 5 sits at the top. Press Unfold and watch the two branches separate downward. Press Solve to fill in the answers. Reset puts it back together.

The widget shows the algebraic recipe in motion. Notice that the two branches are not different problems — they are *the same problem*, broken into the two cases that the absolute value bars were hiding.

The widget hides one subtlety. The "-5" on the right of the second branch did not come out of nowhere. It came from the fact that |y| = 5 has two real-number solutions: y = 5 and y = -5. So if you set y = x - 3, then |x - 3| = 5 forces x - 3 to be either 5 or -5. Why: the absolute value strips the sign. Two distinct real numbers — +5 and -5 — are stripped down to the same value 5. Going backwards, you do not know which one was the "real" value, so you must consider both.

Three worked examples

The recipe is identical every time: spot the bars, write the two branches, solve each branch, collect the answers, check.

Example 1: Solve $|x - 3| = 5$

This is the example from the widget. Watch the unfolding happen on paper.

Step 1. Read k and check it is positive.

k = 5 > 0. So two branches will appear.

Step 2. Unfold into the two branches.

x - 3 = 5 \quad \text{or} \quad x - 3 = -5

Why: |x - 3| = 5 means the expression x - 3 has absolute value 5. The only real numbers with absolute value 5 are +5 and -5, so x - 3 must equal one of those.

Step 3. Solve each branch.

Left branch: x - 3 = 5 \implies x = 8. Right branch: x - 3 = -5 \implies x = -2.

Step 4. Check both candidates in the original equation.

|8 - 3| = |5| = 5. Correct. |-2 - 3| = |-5| = 5. Correct.

Result. x = 8 or x = -2.

Two branches of the equation absolute value of x minus 3 equals 5A diagram showing the original equation absolute value of x minus 3 equals 5 at the top, splitting into two branches below. The left branch reads x minus 3 equals 5 and below it x equals 8. The right branch reads x minus 3 equals negative 5 and below it x equals negative 2. |x − 3| = 5 x − 3 = 5 x − 3 = −5 x = 8 x = −2
Example 1 unfolded on paper. Two branches, two solutions, both valid.

Example 2: Solve $|2x + 1| = 9$

A coefficient on x does not change the recipe at all — the bars come off the same way.

Step 1. k = 9 > 0, so two branches.

Step 2. Unfold.

2x + 1 = 9 \quad \text{or} \quad 2x + 1 = -9

Why: the expression inside the bars is 2x + 1. That entire expression — coefficients, constants, all of it — is what equals \pm 9. The bars do not care about the structure inside, only about the final value.

Step 3. Solve each branch.

Left branch: 2x + 1 = 9 \implies 2x = 8 \implies x = 4. Right branch: 2x + 1 = -9 \implies 2x = -10 \implies x = -5.

Step 4. Check.

|2(4) + 1| = |9| = 9. Correct. |2(-5) + 1| = |-9| = 9. Correct.

Result. x = 4 or x = -5.

Two branches of the equation absolute value of 2x plus 1 equals 9A diagram showing the equation absolute value of 2x plus 1 equals 9 at the top, splitting into two branches: 2x plus 1 equals 9 leading to x equals 4, and 2x plus 1 equals negative 9 leading to x equals negative 5. |2x + 1| = 9 2x + 1 = 9 2x + 1 = −9 x = 4 x = −5
Example 2 unfolded. The coefficient on $x$ inside the bars does not change the unfolding rule — both branches are still ordinary linear equations.

Example 3: Solve $|x^2 - 4| = 5$

Now the expression inside is quadratic. The unfolding rule still works — but you have to be ready for one branch to give zero real solutions.

Step 1. k = 5 > 0, so two branches.

Step 2. Unfold.

x^2 - 4 = 5 \quad \text{or} \quad x^2 - 4 = -5

Why: the bars do not know whether the expression inside is linear, quadratic, or anything else. They just say "the value of this expression has absolute value 5", and the unfolding move applies the same way as before.

Step 3. Solve each branch.

Left branch: x^2 - 4 = 5 \implies x^2 = 9 \implies x = 3 or x = -3. Right branch: x^2 - 4 = -5 \implies x^2 = -1.

That right branch has no real solution — no real number squared gives a negative answer. Why: x^2 \ge 0 for every real x. The right branch is asking for an x whose square is -1, which is impossible over the real numbers. (In CBSE Class 11 onwards, you will meet complex numbers where x = \pm i does satisfy this — but for real-number absolute value equations, this branch contributes nothing.)

Step 4. Check the surviving candidates.

|3^2 - 4| = |9 - 4| = |5| = 5. Correct. |(-3)^2 - 4| = |9 - 4| = |5| = 5. Correct.

Result. x = 3 or x = -3. Total: 2 real solutions (the right branch was empty).

Two branches of the equation absolute value of x squared minus 4 equals 5A diagram showing the equation absolute value of x squared minus four equals five at the top, splitting into two branches. The left branch reads x squared minus four equals five, leading to x squared equals nine, leading to x equals plus or minus three. The right branch reads x squared minus four equals negative five, leading to x squared equals negative one, leading to no real solution. |x² − 4| = 5 x² − 4 = 5 x² − 4 = −5 x² = 9 x² = −1 x = ±3 no real solution
Example 3 unfolded. The left branch contributes two real solutions ($x = \pm 3$). The right branch contributes none — $x^2 = -1$ has no real-number answer. So the original equation has 2 real solutions in total, not 4.

The lesson from Example 3: always unfold first, then check whether each branch is solvable in the real numbers. Quadratic, cubic, or rational expressions inside the bars sometimes give branches with no real solution. That is fine — you just discard those branches and keep what survives.

What about k = 0 and k < 0?

Two edge cases worth saying clearly.

k = 0. The equation |\text{expr}| = 0 unfolds into \text{expr} = 0 and \text{expr} = -0. But -0 = 0, so the two branches collapse into the same equation. You solve \text{expr} = 0 once and you are done. Why: zero is its own negative. The two arrows of the geometric picture have collapsed onto the centre, and algebraically the two branches have collapsed into one.

k < 0. The equation |\text{expr}| = -3 has no solutions at all. You do not even need to unfold — absolute values are non-negative by definition, so a negative k on the right side is impossible to match. The bars cannot equal -3 for any value of the expression inside. Write "no real solution" and move on.

This is one of the few "free" results in algebra: spotting k < 0 saves you all the work.

Why this is the standard CBSE Class 11 technique

NCERT's Class 11 chapter on inequalities and absolute values introduces this two-case unfolding as the primary algebraic tool — before any geometric picture, before any squaring tricks. The reason is that the unfolding rule is mechanical and universal: it works for linear interiors, quadratic interiors, rational interiors, anything. You never have to think "what kind of expression is this?" — you just unfold, then deal with each branch using whatever method that branch needs (linear equation moves, quadratic formula, factoring).

The same recipe carries straight into JEE Advanced problems where the interior is something like |x^2 - 5x + 6| or |\sin x - 1/2|. The bars unfold the same way; only the work inside each branch gets harder. Master the unfolding move on linear interiors, and you have built the muscle that all later absolute value problems will use.

A common slip to avoid

Some students unfold |2x + 1| = 9 as

2x + 1 = 9 \quad \text{or} \quad -2x - 1 = 9

— flipping the sign of the whole left side instead of just the right side. This gives x = 4 and x = -5, which happen to be correct here, but the second branch is doing extra (and wrong) work: you have written -(2x + 1) = 9, which is the same as 2x + 1 = -9. The correct framing is the expression inside the bars equals +k or -k, not "the expression inside, with all signs flipped, equals +k".

Both approaches happen to give the same answers in linear cases, but in equations with multiple absolute values or non-linear interiors, mixing up which side the negative goes on creates real bugs. Stick to the clean rule: \text{expr} = k or \text{expr} = -k, with the negative sitting on the right side both times.

Where this connects

References

  1. NCERT, Mathematics Textbook for Class 11, Chapter 6 (Linear Inequalities) — modulus function and case-based solutions. NCERT Class 11 Mathematics.
  2. Khan Academy — Solving absolute value equations.
  3. Paul's Online Math Notes — Absolute value equations.
  4. Wikipedia — Absolute value.
  5. Art of Problem Solving — Absolute value.