Two numbers at every distance
The absolute value |x| is the distance of x from zero. So |x| = 5 asks for the numbers that are 5 away from zero. There are two of them, one on each side: 5 and −5. The equation |x| = 5 has two solutions, x = 5 and x = −5.
A hop of 5 from zero each way lands on 5 and on −5, the two solutions of |x| = 5.
Split into two cases
Now solve |x − 3| = 5. The bars are around x − 3, so it is x − 3 that is 5 away from zero. That means x − 3 is either 5 or −5, and the bars hide which one. Write both cases and solve each.
If x − 3 = 5, add 3 to both sides: x = 8. If x − 3 = −5, add 3 to both sides: x = −2. The equation has two solutions, x = 8 and x = −2.
Read it as a distance
The same answer comes from the number line. |x − 3| is the distance between x and 3, so |x − 3| = 5 asks for the numbers that are 5 away from 3. Step 5 to the right of 3 to reach 3 + 5 = 8, and 5 to the left to reach 3 − 5 = −2.
So the two solutions always sit the same distance either side of the number in the bars. 3 is halfway between −2 and 8.
8 and −2 are each 5 away from 3, so both solve |x − 3| = 5.
Check both solutions
Put each solution back into the equation. For x = 8: |8 − 3| = |5| = 5. For x = −2: |−2 − 3| = |−5| = 5. The inside comes out as 5 once and as −5 once, and the bars give 5 both times.
A plus sign inside the bars
Solve |x + 2| = 6. The two cases are x + 2 = 6, which gives x = 4, and x + 2 = −6, which gives x = −8.
As a distance, x + 2 is x − (−2), so |x + 2| is the distance between x and −2. The solutions are 6 either side of −2: −2 + 6 = 4 and −2 − 6 = −8.
The same split works when x is multiplied by a number. For |2x − 1| = 7, either 2x − 1 = 7, so 2x = 8 and x = 4, or 2x − 1 = −7, so 2x = −6 and x = −3.
|x + 2| = 6: the solutions 4 and −8 are each 6 away from −2.
Two solutions, one, or none
A distance is never negative, so an absolute value is never below zero. |x + 1| = −4 asks for a distance of −4, so the equation has no solution.
When the right side is zero there is only one solution. |x − 3| = 0 asks for the numbers that are no distance from 3, and the only one is x = 3.
So look at the right side before splitting. If it is positive there are two solutions, if it is zero there is one, and if it is negative there are none.
The usual mistakes
Solving only one case. x − 3 = 5 gives x = 8, but x − 3 = −5 is just as possible, and it gives x = −2. An answer with one solution has missed the other.
Changing the sign of the answer instead of the inside. |x − 3| = 5 does not give x = 8 or x = −8. The bars are around x − 3, so it is x − 3 that can be negative: |−8 − 3| = 11, not 5.
Splitting when the right side is negative. |x + 1| = −4 split into x + 1 = −4 or x + 1 = 4 gives x = −5 and x = 3, and neither works: |−5 + 1| = 4 and |3 + 1| = 4, not −4.
Worked example: The Two Alarm Temperatures of a Medicine Fridge
Question A pharmacy keeps a medicine fridge at 5 °C. Its alarm is set to sound at the two temperatures that are exactly 3 °C away from 5 °C. (a) Write an absolute value equation for the alarm temperatures, x °C, and solve it. (b) The pharmacy's freezer has its alarm set to sound at −25 °C and at −15 °C. Write these two temperatures as the solutions of one equation |x − a| = b, giving the values of a and b.
1.The distance between x and 5 on the temperature scale is |x − 5|. The alarm sounds when this distance is 3, so |x − 5| = 3.
The distance between x and 5 is |x − 5|, and the alarm sounds when it is 3: |x − 5| = 3. 2.The number inside the bars is either 3 or −3: x − 5 = 3 or x − 5 = −3.
The number inside the bars is 3 or −3: x − 5 = 3 or x − 5 = −3. 3.Add 5 to both sides of each: x = 8 or x = 2. (a) |x − 5| = 3, and the alarm sounds at 2 °C and at 8 °C. Check: |8 − 5| = 3 and |2 − 5| = |−3| = 3.
(a) Add 5 to both sides of each: x = 8 or x = 2. The alarm sounds at 2 °C and at 8 °C. 4.The solutions of |x − a| = b are a − b and a + b, so a is halfway between them: a = −25 + (−15)2 = −402 = −20.
The solutions of |x − a| = b are a − b and a + b, so a is halfway between −25 and −15: a = −20. 5.b is half the gap between them: b = −15 − (−25)2 = 102 = 5. (b) a = −20 and b = 5, so the equation is |x − (−20)| = 5, that is |x + 20| = 5. Check: |−25 + 20| = 5 and |−15 + 20| = 5.
(b) b is half the gap of 10, so b = 5, and the equation is |x + 20| = 5.
Answer: (a) |x − 5| = 3, so x = 2 or x = 8: the alarm sounds at 2 °C and at 8 °C; (b) a = −20 and b = 5, so |x + 20| = 5
Common mistakes
- Solving only x − 5 = 3 and giving the single answer 8 °C. The bars hide the sign of x − 5: a temperature 3 °C below 5 °C is also 3 °C away, so x = 2 is a second solution.
- Writing the freezer's equation as |x − 20| = 5. Its solutions are 15 and 25, both above freezing. The middle of −25 and −15 is −20, and x − (−20) is x + 20.
More equations and inequalities problems, worked step by step →