2
Substitution set and solution set: finding solutions by testing values
Mathematics Prep 1 Term 1, Unit 2, Lesson 2
GoalFind the solution set of an equation inside a given substitution set by testing its numbers.
New words
substitution set · the set of numbers we are allowed to try for xsolution set · the numbers of the substitution set that make the equation truesolution · a number that makes the equation trueempty set · ∅: the solution set when no number works
Explanation
An equation is only a question until we say which numbers we may use. The substitution set is that set of allowed numbers, such as N, Z, Q or a list like {1, 2, 3, 4, 5}. To solve by testing, put each number into the equation. The numbers that make it true form the solution set. The solution set is always a subset of the substitution set. The same equation can have different solution sets in different substitution sets: x + 2 = 0 has the solution −2 in Z, but no solution in N, so its solution set in N is ∅.
★Remember
Solution set ⊂ substitution set. Test every number. If none works, the solution set is ∅.
Worked examples
| x | x + 3 | Equals 7? |
|---|---|---|
| 1 | 4 | no |
| 2 | 5 | no |
| 3 | 6 | no |
| 4 | 7 | yes |
| 5 | 8 | no |
Solve 2x − 1 = 9 in {2, 3, 4, 5, 6}. For x = 5: 2 × 5 − 1 = 9, which is true. The other numbers fail. The solution set is {5}. In the substitution set {1, 2, 3} no number works, so the solution set is ∅.
The same equation in different sets. x + 2 = 0. In N the answer −2 is not allowed, so the solution set is ∅. In Z the solution set is {−2}. And 3x = 7 has no solution in Z, but in Q the solution set is {7/3}.
Class activity
Try it out. With a grown-up, write the equation x + 4 = 9 and the set {3, 4, 5, 6}. Test each number aloud. Then change the equation to x + 4 = 2 and test again in N, then in Z.
Practice
1
Solve x + 6 = 10 in {2, 3, 4, 5}.
Only 4 works: {4}.
2
Solve 3x = 12 in {1, 2, 3}.
3 × 1 = 3, 3 × 2 = 6, 3 × 3 = 9. None gives 12, so the solution set is ∅.
3
Solve x + 5 = 2 in N and in Z.
x = −3. In N the solution set is ∅. In Z it is {−3}.
4
Solve 2x + 1 = 11 in {3, 4, 5, 6}.
2 × 5 + 1 = 11, so the solution set is {5}.
5
Hana says the solution set of x + 2 = 5 in {1, 2} is {3}. What is wrong?
3 is not in the substitution set {1, 2}. Neither 1 nor 2 works, so the solution set is ∅.