Absolute value equations

When you reach for it

The absolute value |x| is how far x sits from 0 on the number line, with no regard for which side. So the equation

asks: which numbers are 5 away from 0? There are two, one on each side: x = 5 and x = -5. Both make the sentence true, since |5| = 5 and |-5| = 5. The pair is often written x = \pm 5.

Reach for this move whenever the unknown sits inside absolute value bars: a score that is a certain number of points from a target, a loaf baked a certain number of minutes longer or shorter than the recipe says, a point a given distance from another point. Each of these states how far, and leaves out which way, so each has two candidate answers, one for each direction.

The procedure

Why there are two cases

Absolute value has a two-part rule: |u| = u when u \ge 0, and |u| = -u when u < 0. (The absolute value graph draws that rule as a V.) Suppose d is a positive number and |u| = d. Either u \ge 0, and then u = d; or u < 0, and then -u = d, which means u = -d. Going back the other way, both u = d and u = -d do have absolute value d. So, for any expression u and any number d > 0,

The absolute value equation becomes two ordinary linear equations, one for each side of the number line.

The size of d decides how many solutions there are:

• If d > 0, there are two cases, as above.
• If d = 0, the two cases are the same equation, u = 0, because -0 = 0. Only one number is at distance 0 from zero, namely zero itself.
• If d < 0, there is no solution. A distance is never negative, so |u| is never below 0 and cannot equal a negative number. Nothing needs to be solved: the right-hand side settles it.

Isolate the absolute value first

The rule above is about |u| standing alone on one side. In

the bars are multiplied by 2 and then have 3 added, so the rule does not apply yet. Undo those two operations, the same way as in any two-step equation, treating |x - 1| as a single unknown quantity. The properties of equality allow it: subtract 3 from both sides, then divide both sides by 2:

Now the bars stand alone and the right side is positive, so split into the two cases:

which give x = 5 or x = -3.

Reading |x - a| = d as a distance

The expression |x - a| is the distance between x and a on the number line. So |x - 1| = 4 asks for the points at distance 4 from 1: one is 4 to the right, at 5, and one is 4 to the left, at -3. That is the same pair the algebra found.

The picture also shows where the solutions lie: always symmetric about a, which is their midpoint. A sum inside the bars, such as |x + 2|, measures distance from a negative number: because x + 2 = x - (-2), |x + 2| is the distance from x to -2.

Checking both answers

A solution is a number that makes the original sentence true, so check each one by putting it back into the equation you started with, not into a later line:

Both are true. Inside the bars, the two answers give opposite numbers, 4 and -4, and the bars turn both into 4. That is exactly why both work.

Worked examples

Boundaries

• The two-case rule |u| = d \iff u = d or u = -d needs d \ge 0. When d < 0 there is no solution, and when d = 0 the two cases coincide.
• It applies only once the absolute value stands alone. A number multiplying the bars or added to them must be moved first, as in any linear equation.
• d must be a known number. If the unknown also appears outside the bars, as in |x| = x - 2, the sign of the right-hand side is unknown, a case can produce a number that fails, and checking in the original stops being a safeguard and becomes part of the solution (exercise 17).
• Inside the bars the expression here is linear. With x^2 inside, each case is a quadratic equation, and there may be more than two solutions.

Where this goes next

Change the equals sign to < or > and the two edge points become the ends of a stretch of the number line: absolute value inequalities find every point within, or beyond, a given distance. On the graph, the solutions of |x - a| = d are where the horizontal line y = d cuts the V of y = |x - a|, which is how transforming the absolute value graph reads the same equation. With absolute value on both sides, |x - 1| = |x - 7| asks for the point equally far from 1 and 7, which is the midpoint, x = 4. And the pattern of two cases returns in the square root property: x^2 = 9 gives x = \pm 3 for the same reason, since \displaystyle \sqrt{x^2} = |x|.

Exercises

Read the right-hand side

1. Solve |a| = 14.
2. Solve |a + 9| = 0.
3. Solve |a| = -0.5.
4. Solve |c - 7| = 11.

Isolate, then split

5. Solve |x| - 6 = 10.
6. Solve 4|x - 5| = 28.
7. Solve 2|x + 6| + 1 = 15.
8. Solve |3x - 4| = 11.
9. Solve |5 - 2x| = 3.
10. Solve 6 - |x + 1| = -3.
11. Solve \displaystyle \left|\frac{x}{2} - 1\right| + 4 = 7.
12. Solve 3|2x + 1| + 10 = 4.

Distance on the number line

13. Write an absolute value equation whose solutions are the numbers at distance 2.5 from 6, and solve it.
14. Find an equation of the form |x - a| = d whose solutions are -7 and 3.

Problems

15. A class sets a target of 80 points on a quiz. Mabli’s score is 6 points away from the target. Write an absolute value equation for her score s and find the two possible scores.
16. A straight footpath has distance markers in meters. A bench stands at the 120 m marker, and Imelda is standing 35 m from the bench, on the path. Write and solve an equation for her marker p.

Reasoning

17. Solve |x| = x - 2 by splitting into the cases x \ge 0 and x < 0, and check what you find. Explain why the equation has no solution.
18. Explain why |x - 3| = -2 has no solution while |x| - 3 = -2 has two, and find them.