Solving absolute value equations

TL;DR:

• The absolute value |a| is the distance of a from 0, so it is never negative.

• |expression| = k with k positive splits into two equations: expression = k and expression = −k.

• Always check your answers, because squeezing a variable on both sides can give extraneous solutions.

The temperature is 4 degrees away from freezing. Is it 4 above or 4 below? Both fit the sentence. That two-way ambiguity is the heart of absolute value, and it is why equations with an absolute value usually have two solutions.

Absolute value as a distance

The absolute value of a number is its distance from zero on the number line. So |5| = 5 and |−5| = 5. Written as a rule: |a| = a if a is zero or positive, and |a| = −a if a is negative. The second line does not mean the result is negative, since the negative of a negative number is positive.

Generalising, |x − a| is the distance between x and a. The equation |x − 2| = 3 asks for the points 3 units from 2, which are 5 and −1.

Worked example: |2x − 3| = 7

The thing inside the bars is either 7 or −7, because both have absolute value 7. Write down both cases.

1. Case 1: 2x − 3 = 7. Add 3: 2x = 10. Divide by 2: x = 5.

2. Case 2: 2x − 3 = −7. Add 3: 2x = −4. Divide by 2: x = −2.

Check both in the original equation. For x = 5: |2(5) − 3| = |7| = 7. For x = −2: |2(−2) − 3| = |−7| = 7. Both work, so the solutions are x = 5 and x = −2.

First isolate the absolute value

If other terms surround it, move them away before splitting. For |2x − 3| + 4 = 11, subtract 4: |2x − 3| = 7, the same equation as above. If you split before isolating, you will get wrong cases.

When there is no solution

A distance cannot be negative. So |x + 1| = −2 has no solution. Always look at the right-hand side first: if it is negative, stop. If it is zero, there is just one case, since |u| = 0 means u = 0.

Extraneous solutions

When the right-hand side contains x, the cases can produce answers that fail the check. Solve |x − 1| = 2x − 4.

1. Case 1: x − 1 = 2x − 4, so 3 = x. Check: left |3 − 1| = 2, right 2(3) − 4 = 2. It works.

2. Case 2: x − 1 = −(2x − 4) = −2x + 4, so 3x = 5 and x = 5/3. Check: left |5/3 − 1| = 2/3, right 2(5/3) − 4 = −2/3. The sides differ, so 5/3 is extraneous.

The only solution is x = 3. The second case treated the right side as if it could be negative. But an absolute value is never negative, and at x = 5/3 the right side is −2/3. The check catches it.

Try two more yourself

Solve |3x + 2| = 10. Case 1: 3x + 2 = 10, so 3x = 8 and x = 8/3. Case 2: 3x + 2 = −10, so 3x = −12 and x = −4. Check: |3(8/3) + 2| = |10| = 10, and |3(−4) + 2| = |−10| = 10. Both are valid.

Now one that needs isolating first: 2|x − 1| − 5 = 3. Add 5: 2|x − 1| = 8. Divide by 2: |x − 1| = 4. So x − 1 = 4 or x − 1 = −4, which gives x = 5 or x = −3. Check x = −3: 2|−4| − 5 = 8 − 5 = 3. Check x = 5: 2|4| − 5 = 3. Both work. The answers are 4 either side of 1, which is what the distance reading predicts.

Inequalities in a sentence

The same distance idea solves inequalities. |x − 2| < 3 means x is within 3 of 2: −1 < x < 5. And |x − 2| > 3 means x is more than 3 away: x < −1 or x > 5. Check one point each: x = 0 gives |−2| = 2 < 3, inside. x = 6 gives |4| = 4 > 3, outside.

A picture helps. The graph of y = |x| is a V with its point at the origin. Solving |x| = k means drawing the horizontal line y = k and reading where it meets the V. For positive k the line cuts the V twice, at −k and k. For k = 0 it touches just the point. For negative k it never meets the V, which is the no-solution case from above. The same picture, shifted and stretched, works for |2x − 3| = 7.

How to check your answer

• Substitute each answer into the original equation.

• Number line. For |x − a| = k, the answers are a + k and a − k, equally far from a. Here a = 3/2 and k = 7/2: 3/2 + 7/2 = 5, 3/2 − 7/2 = −2.

• Sign of the right side. It must be zero or positive.

• Count. Two solutions are typical; one or none are special cases.

Common mistakes

• Writing only the positive case.

• Splitting before isolating the absolute value.

• Forgetting to flip all signs in the negative case: −(2x − 4) is −2x + 4, not −2x − 4.

• Skipping the check.

• Dropping the bars and treating |a| as a.

Where you will use it next

Absolute value links to the sign rules in inequalities and why the sign flips and to distance in graphs. Counting cases carefully is a skill shared with permutations and combinations.

Frequently asked questions

Why are there two answers?

Two different numbers can have the same distance from zero. Splitting into cases finds both.

Can an absolute value equation have three solutions?

Not a simple one like these. With more bars, such as ||x| − 2| = 1, up to four solutions can occur.