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Questions like |x − 3| + |x + 1| = 6 or |x − 3| > 2x look short, and most of the marks are lost in the same place: an answer that solves the algebra but lies outside the region where that algebra was valid. This post gives one method that handles equations and inequalities alike, works five examples, and shows the slips to avoid.
Where is this in the syllabus?
The NCERT Class 11 chapter Relations and Functions (2026-27 reprint) defines the modulus function as f(x) = x for x ≥ 0 and −x for x < 0. The chapter Linear Inequalities covers algebraic solutions of linear inequalities in one variable. The JEE (Main) 2026 syllabus has no separate unit on inequalities, but these skills sit inside its units on functions, quadratic equations and calculus. The JEE (Advanced) 2026 brochure names the "absolute value" function among its special functions. Check your year's syllabus.
What does the research say about these mistakes?
Almog and Ilany (2012) gave a questionnaire on absolute value inequalities to 481 students in grades 10 and 11 in Israel, after they had studied inequalities, and interviewed 32 of them. They found main types of mistakes that students made consistently, and traced them to the students' ways of thinking. In a separate teaching study, Sierpinska, Bobos and Pruncut (2011) found that students taught both a graphical and an analytic technique, each usable to check the other, dealt more effectively with the logic of these inequalities than groups taught a procedural or a theoretical approach, though the authors note the graphical route also let some students avoid the harder analytic thinking.
That suggests a habit: solve by cases, then check with a quick sketch or a test value.
What is the case-split method?
Find the critical points, where each expression inside a modulus equals zero.
Split the number line at those points.
On each piece, replace each |E| by E or −E, depending on the sign of E there.
Solve, then keep only solutions inside that piece.
Combine the pieces, and check one value.
Worked example 1: two moduli
Solve |x − 3| + |x + 1| = 6. Critical points: −1 and 3.
x < −1: both negative inside, so −(x − 3) − (x + 1) = 6, giving −2x + 2 = 6 and x = −2. It is less than −1, so keep it.
−1 ≤ x ≤ 3: (3 − x) + (x + 1) = 4. The equation says 4 = 6, so no solution here.
x > 3: (x − 3) + (x + 1) = 6, so 2x = 8 and x = 4. Keep it.
Answer: x = −2 or x = 4. Check: |−5| + |−1| = 6 and |1| + |5| = 6.
The middle piece shows a picture worth remembering: |x − 3| + |x + 1| is the sum of the distances from x to 3 and to −1. Between them that sum is always 4, the gap between the points.
Worked example 2: a variable on the right
Solve |x − 4| = 2x + 1.
x ≥ 4: x − 4 = 2x + 1 gives x = −5. Not in x ≥ 4, so reject it.
x < 4: 4 − x = 2x + 1 gives x = 1. It is in x < 4, so keep it.
Answer: x = 1. Check: |1 − 4| = 3 and 2(1) + 1 = 3. Students who square both sides get x = 1 and x = −5, and −5 fails because 2(−5) + 1 is negative while a modulus never is.
Worked example 3: an inequality
Solve |x − 1| + |x − 4| ≤ 5. Critical points: 1 and 4.
x < 1: (1 − x) + (4 − x) = 5 − 2x ≤ 5, so x ≥ 0. With x < 1: 0 ≤ x < 1.
1 ≤ x ≤ 4: (x − 1) + (4 − x) = 3 ≤ 5, true for the whole piece.
x > 4: 2x − 5 ≤ 5, so x ≤ 5. With x > 4: 4 < x ≤ 5.
Combine: 0 ≤ x ≤ 5. Check x = 5: 4 + 1 = 5. Check x = 6: 5 + 2 = 7, which is too big.
Worked example 4: inequality with x on both sides
Solve |x − 3| > 2x.
x ≥ 3: x − 3 > 2x gives x < −3, which cannot happen when x ≥ 3. Nothing here.
x < 3: 3 − x > 2x gives x < 1. Combined with x < 3: x < 1.
Answer: x < 1. Check x = 0: 3 > 0, true. Check x = 1: 2 > 2, false, so 1 is excluded.
Worked example 5: a quadratic in |x|
Solve x² − 5|x| + 6 = 0. Since x² = |x|², put u = |x|: u² − 5u + 6 = 0, so u = 2 or 3. Then x = ±2 or ±3, four solutions. No case split is needed when every x appears as |x| or x².
Common mistakes and how to avoid them
Not rejecting answers outside their piece. Write the piece's condition next to every answer.
Treating |x − 2| < 3 like |x − 2| > 3. The first is one interval, −1 < x < 5. The second is two, x < −1 or x > 5.
Squaring when the other side can be negative. Squaring is safe only when both sides are known to be non-negative.
Losing endpoints. Decide for each critical point which piece it belongs to, and test it in the original.
Our post on the wavy curve method handles rational inequalities, and graph transformations shows how to sketch y = |f(x)| for a visual check.
A practice routine
Day 1: Five one-modulus equations, each with a variable on the right.
Day 2: Three two-modulus equations; sketch the left side as a broken line to check.
Day 3: Four inequalities, two of each direction.
Day 4: Mixed set of six, timed at three minutes each.
If you get stuck choosing the pieces, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Tonight, solve |2x − 1| = x + 2 by cases and check that both x = 3 and x = −1/3 survive.
