Guides / Equations

Absolute value equations: two cases, then check

An absolute value equation is two ordinary equations in one. The points get lost in three places: splitting before the bars are alone, negating only part of the other side, and skipping the check.

Why it works

|u| is the distance from u to 0 on the number line. Distances are never negative, and there are exactly two numbers at distance 7 from 0: 7 and −7. So |u| = 7 means u = 7 or u = −7, and nothing else. That's the whole method. Everything below is about applying it to the right u, and to the right 7.

Example 1: the basic split

Solve |2x − 3| = 7.

1. 2x − 3 = 7 or 2x − 3 = −7The absolute value is already alone and 7 is positive, so split.
2. 2x = 10 or 2x = −4Add 3 to both sides of each.
3. x = 5 or x = −2Divide by 2.

Check: |2(5) − 3| = |7| = 7 and |2(−2) − 3| = |−7| = 7. Both work.

Example 2: isolate the absolute value first

Solve 3|x + 4| − 5 = 13.

1. 3|x + 4| = 18Add 5. Treat |x + 4| like a single block you're solving for.
2. |x + 4| = 6Divide by 3. Now the bars are alone, so it's time to split.
3. x + 4 = 6 or x + 4 = −6Two cases.
4. x = 2 or x = −10Subtract 4 from each.

Check: 3|6| − 5 = 13 and 3|−6| − 5 = 13.

Example 3: x on both sides, and an answer that fails

Solve |x − 5| = 2x + 1.

1. x − 5 = 2x + 1Case 1: the inside equals the other side.
2. x = −6Subtract x and 1 from both sides: −6 = x.
3. x − 5 = −(2x + 1)Case 2: the inside equals the opposite of the whole other side. Keep the parentheses.
4. x − 5 = −2x − 1Distribute the minus to both terms.
5. x = 43Add 2x and 5: 3x = 4.
6. x = 43 onlyCheck both. x = −6: left side |−11| = 11, right side 2(−6) + 1 = −11. 11 ≠ −11, so −6 is rejected. x = 4/3: left |4/3 − 5| = 11/3, right 8/3 + 1 = 11/3. It works.

Why did x = −6 show up at all? Case 1 assumed the right side equals the inside, but at x = −6 the right side is negative, and an absolute value can't equal a negative number. The algebra in each case is fine; the check is what filters the answers. Whenever the right side has an x in it, the check is part of the solution, not an optional extra.

Common mistakes, and the exact line they happen on

1. Splitting before the absolute value is alone

Solve 3|x + 4| − 5 = 13 (Example 2).

1. 3(x + 4) − 5 = −13This is the mistake. The −13 applies the "negative case" to the whole left side, including the −5 and the 3 that are outside the bars. This line leads to x = −20/3, which doesn't satisfy the original. Isolate first: |x + 4| = 6, then x + 4 = −6, so x = −10.

Only the expression inside the bars gets the ±. Anything outside them has to be moved away first.

2. Negating only the first term of the other side

Solve |x − 5| = 2x + 1 (Example 3).

1. x − 5 = −2x + 1This is the mistake. The opposite of 2x + 1 is −2x − 1, not −2x + 1. This line gives x = 2, and the check catches it: |2 − 5| = 3, but 2(2) + 1 = 5.

Write the second case as −( … ) with parentheses around the entire other side, then distribute on the next line. It's the same dropped-parenthesis error that shows up when substituting in a system of equations.

3. Solving an equation that has no solution

1. |x + 2| = −4The absolute value is alone, and it equals a negative number.
2. x + 2 = −4 or x + 2 = 4This is the mistake. Splitting gives x = −6 or x = 2, and both make |x + 2| equal 4, not −4. An absolute value is never negative, so the answer is no solution. Stop at the line above.

4. Skipping the check when x is on both sides

In Example 3, writing "x = −6 or x = 4/3" as the final answer loses the point even though every line of algebra was right. When both sides contain x, substitute each answer into the original equation. When the other side is a plain positive number, as in Examples 1 and 2, both cases always work and the check is just insurance.

5. Treating the bars like parentheses

|x − 3| is not x + 3, and −|x − 3| is not |−x + 3|. Absolute value doesn't distribute and doesn't absorb a sign from outside. If a minus sign sits in front of the bars, move it with the rest of the outside material while isolating: −|x − 3| = −5 becomes |x − 3| = 5.

Practice

1. Solve |x − 4| = 9.

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   x − 4 = 9 or x − 4 = −9, so x = 13 or x = −5.

2. Solve 2|x − 1| + 3 = 11.

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   Isolate: 2|x − 1| = 8, |x − 1| = 4. Then x − 1 = 4 or x − 1 = −4, so x = 5 or x = −3.

3. Solve |x + 6| + 8 = 3.

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   Isolate: |x + 6| = −5. An absolute value can't be negative, so there is no solution.

4. Solve |2x − 1| = x + 4.

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   Case 1: 2x − 1 = x + 4, x = 5. Case 2: 2x − 1 = −(x + 4) = −x − 4, 3x = −3, x = −1. Check: |9| = 9 ✓ and |−3| = 3 = −1 + 4 ✓. Both work: x = 5 or x = −1.

5. Solve |x + 3| = 2x.

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   Case 1: x + 3 = 2x, x = 3. Case 2: x + 3 = −2x, x = −1. Check: |6| = 6 = 2(3) ✓, but |2| = 2 while 2(−1) = −2 ✗. Only x = 3.