An answer that is many numbers
The equation 2x + 1 = 7 has one solution. Subtract 1 from both sides to get 2x = 6, then divide both sides by 2 to get x = 3.
Now solve the inequality 2x + 1 > 7 with the same moves. Subtracting 1 gives 2x > 6, and dividing by 2, which is positive, keeps the sign as it is: x > 3. This answer is not one number. 4 works, since 2 × 4 + 1 = 9 is more than 7. So do 3.5, 10 and 1000. Every number greater than 3 works, and all of them together make up the solution set of the inequality.
The number 3 itself is not in the set: 2 × 3 + 1 = 7, and 7 is not more than 7. The set starts just after 3 and goes on without end.
The solution set of 2x + 1 > 7. The circle at 3 is hollow because 3 is not in the set, and the arrow shows that the set carries on past 8.
Writing the set down
A set with only a few members can be listed inside curly brackets, like {1, 2, 3}. The solution set of x > 3 cannot be listed, because it has infinitely many members: 3.1, 3.01, 3.001 and every other number above 3. So the set is described by the rule its members follow instead.
Set-builder notation does this. The solution set of x > 3 is written {x : x > 3}. The curly brackets mean "the set of", the x before the colon names the members, and the colon means "such that". After the colon comes the condition that every member must satisfy. Read aloud, {x : x > 3} is "the set of all x such that x is greater than 3".
Some books write a vertical bar in place of the colon, as {x | x > 3}. It means exactly the same thing.
Keeping the endpoint
The sign means "greater than or equal to". So {x : } is every number greater than −1, together with −1 itself. Check the endpoint by putting it into the condition: is true, because −1 is equal to −1.
On the number line, an endpoint that belongs to the set is drawn as a filled dot, and an endpoint that does not belong is drawn as a hollow circle. The dot is the only difference between the drawings of and x > −1.
The set {x : }. The dot at −1 is filled because −1 is a member of the set.
A set with two ends
A set can be bounded on both sides. {x : } is every number that is greater than −1 and also less than or equal to 4. Both conditions must hold at once, so the set is the stretch of the number line between −1 and 4.
Read each end from its own sign. The < beside −1 is strict, so −1 is not in the set and its circle is hollow. The beside 4 includes 4, so its dot is filled.
Test a few numbers. 0 and 2.5 are members. 4 is a member, since . −1 is not, since −1 < −1 is false, and 5 is not, since is false.
The set {x : }: hollow at −1, which is left out, and filled at 4, which is included.
The usual mistakes
Writing the answer as {3}. The set {3} has exactly one member, the number 3, and 3 is not even a solution of x > 3. The solution set is {x : x > 3}.
Mixing up the dots. A hollow circle means the endpoint is left out, which goes with < and >. A filled dot means it is included, which goes with and .
Writing the two ends in the wrong order. {x : } asks for numbers greater than 4 and at most −1, and no number is both. Write the smaller end on the left, as in .
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 →