Within a distance of zero
|x| is the distance of x from zero, so |x| < 4 asks for the numbers that are less than 4 away from zero. On the right of zero they run up to 4, and on the left down to −4. Neither 4 nor −4 is included, because each is exactly 4 away, not less. The solution is −4 < x < 4.
It is not x < 4 alone. −10 is less than 4, but it is 10 away from zero, so |−10| = 10 is not less than 4.
The numbers less than 4 from zero: the stretch between −4 and 4, with both ends hollow.
Within a distance of another number
|x − 3| is the distance between x and 3. So |x − 3| < 5 asks for the numbers less than 5 away from 3.
In symbols, the inside, x − 3, is less than 5 away from zero, so it lies between −5 and 5: −5 < x − 3 < 5. Add 3 to all three parts to get x on its own: −2 < x < 8.
On the number line that is one stretch, centered on 3 and reaching 5 each way, from 3 − 5 = −2 up to 3 + 5 = 8.
The solutions of |x − 3| < 5: every number between −2 and 8, which are both 5 away from 3.
The general rule
Every inequality of this kind works the same way. |x − a| < b means that x is less than b away from a, so a − b < x < a + b. The stretch is centered on a, and its width is 2b.
With instead of <, the two ends are exactly b away from a, which is allowed, so they are included: means , drawn with filled circles.
|x − c| ≤ d is every x within distance d of c: the corridor from c − d = 0 to c + d = 2
Set c = 3 and d = 2 to solve |x − 3| ≤ 2
The handle on the line moves the center, c, and the handle below it sets the distance, d. The shaded stretch is every x within d of c, from c − d to c + d. Set c = 3 and d = 2 to solve .
Further away: two rays
Reverse the sign. |x − 3| > 5 asks for the numbers more than 5 away from 3. They are on both sides: past 8 on the right, and past −2 on the left.
In symbols, the inside is more than 5 away from zero, so either x − 3 > 5 or x − 3 < −5. Add 3 to each: x > 8 or x < −2. The answer is written x < −2 or x > 8.
These are two rays heading apart, with a gap between them, and no single double inequality can describe them. Writing 8 < x < −2 would ask for a number greater than 8 and less than −2 at the same time, and there is no such number.
The solutions of |x − 3| > 5: two rays, one running left from −2 and one running right from 8.
Which shape to expect
Less than a distance gives one stretch, and its two ends are joined by "and": x is more than a − b and less than a + b. More than a distance gives two rays, joined by "or": x is below a − b or above a + b.
Check with a value between the ends. x = 3 gives |3 − 3| = 0, which is less than 5, so 3 is in the stretch for |x − 3| < 5 and is not a solution of |x − 3| > 5.
The usual mistakes
Dropping one end. |x − 3| < 5 is not x < 8. A number such as −10 is less than 8, but it is 13 away from 3.
Giving the wrong shape. |x − 3| > 5 is not −2 < x < 8. That stretch is the numbers within 5 of 3, the opposite of what > asks for.
Dropping one ray. x > 8 leaves out the numbers below −2, which are more than 5 away from 3 as well.
Worked example: A Machined Rod Accepted Within a Tolerance
Question A machine cuts steel rods that should be 50 mm long. A rod of length x mm is accepted when its length differs from 50 mm by at most 0.5 mm. (a) Write this condition as an absolute value inequality and solve it. (b) Five rods measure 49.4 mm, 49.5 mm, 50.2 mm, 50.6 mm and 50.5 mm. How many of them are accepted?
1.The distance between x and 50 on the number line is |x − 50|. It must be at most 0.5, so |x − 50| ≤ 0.5.
The distance between x and 50 on the number line is |x − 50|, and it must be at most 0.5. 2.An absolute value is at most 0.5 when the number inside is between −0.5 and 0.5. This gives two inequalities: x − 50 ≥ −0.5 and x − 50 ≤ 0.5.
The number inside is between −0.5 and 0.5: x − 50 ≥ −0.5 and x − 50 ≤ 0.5. 3.Add 50 to both sides of each: x ≥ 49.5 and x ≤ 50.5. (a) |x − 50| ≤ 0.5, and the solution is 49.5 ≤ x ≤ 50.5. On the number line both ends have closed dots, because both end values are accepted.
(a) Add 50 to both sides of each: 49.5 ≤ x ≤ 50.5. Both ends have closed dots. 4.Test each rod. |49.4 − 50| = 0.6 and |50.6 − 50| = 0.6, and these are more than 0.5. |49.5 − 50| = 0.5, |50.2 − 50| = 0.2 and |50.5 − 50| = 0.5, and these are at most 0.5.
Each cross is one rod. The rods of 49.4 mm and 50.6 mm are 0.6 mm from 50 mm. 5.(b) 3 of the five rods are accepted: the rods of 49.5 mm, 50.2 mm and 50.5 mm.
(b) 3 of the five rods are accepted.
Answer: (a) |x − 50| ≤ 0.5, so 49.5 ≤ x ≤ 50.5; (b) 3 rods
Common mistakes
- Writing only x − 50 ≤ 0.5 and accepting every rod shorter than 50.5 mm. A rod can also be too short. The absolute value gives a second inequality, x − 50 ≥ −0.5.
- Rejecting the rods of 49.5 mm and 50.5 mm. The sign is ≤, so a difference of exactly 0.5 mm is accepted and both end values belong to the solution.
More equations and inequalities problems, worked step by step →