045i - Verifying Solutions by Substitution

Lesson 45 — Verifying Solutions to Equations and Inequalities by Substitution

Strand: Algebra | Descriptor: AC9M8A02 | Duration: 45 minutes

Learning Intentions

• To verify solutions to linear equations by substitution.
• To verify whether a value lies within the solution set of an inequality by substitution, including testing boundary values.

Success Criteria

I can:

1. Substitute a value into an equation to check whether LHS RHS.
2. Substitute a value into an inequality to check whether the inequality statement is true.
3. Test a boundary value to decide whether an endpoint is included (closed) or excluded (open).
4. Use substitution to locate an error in equation or inequality solving.

Warmup

(5 minutes — true or false, mini whiteboards)

Decide true or false, showing your substitution.

1. solves .
2. satisfies .
3. satisfies .
4. satisfies .

Answers: 1. True, ; 2. False, … (check: , and is true — so True); 3. True — the boundary is included; 4. False, , and .

Teacher note: correct the discussion answer for Q2 live: is indeed true, so does satisfy it — use this as an opportunity to model checking your own working aloud.

Activities

Activity 1 — Verifying Equations (10 min)

I do. Is the solution to ?

I do — with a fraction. Is the solution to ?

We do:

1. Is the solution to ?
2. Is the solution to ?

(Answers: 1. LHS , RHS ✓ yes; 2. LHS , RHS ✓ yes.)

You do:

1. Is the solution to ? (Answer: LHS , RHS — no.)
2. Is the solution to ? (Answer: LHS , RHS — yes.)

Activity 2 — Verifying Inequalities and Testing Boundaries (10 min)

I do. Does satisfy ?

Yes — the boundary value itself satisfies a inequality.

I do — testing to find the open/closed boundary. For the solution set , does satisfy the original inequality ?

So does not satisfy it — confirming the open boundary. Test : ✓ — just inside.

Key idea: substitution is exactly how you can prove whether a boundary should be open or closed, without relying on memory of the symbol rule.

We do:

1. Does satisfy ?
2. Does satisfy ?

(Answers: 1. , and ✓ yes; 2. , and — no.)

You do:

1. Does satisfy ?
2. Does satisfy ?
3. Does satisfy ?

(Answers: 1. Yes, ; 2. Yes, ; 3. Yes, .)

Activity 3 — Error-hunting across Equations and Inequalities (14 min)

Pairs. Verify first, then locate the exact step where the error occurred.

Student A claims solves , based on this working:

Student B claims the solution set to is .

Student C claims is included in the solution set of .

Socratic scaffolding for Student B:

Prompt	Purpose
First, verify with a test value.	Try : ; is ? Yes. So the boundary checks out — but is the direction right?
Test a value that should be inside the claimed set, e.g. .	; is ? No — this fails, so should NOT be in the true solution set.
So is correct?	No — the true solution set must be on the other side.
Where did the error occur?	Dividing by without reversing the inequality sign.
Redo the step correctly.	(reverse, since dividing by a negative).
Verify the corrected solution with (should now fail) and (should now pass).	: false, correctly excluded. : true, correctly included.

Answers: Student A — verify: LHS , RHS ✓. Student A’s working and claim are actually correct — a deliberate “trick” item to reinforce that not every flagged claim contains an error; verification is unbiased. Student B — should be (sign not reversed when dividing by ). Student C — LHS at : ; is ? No — is not included (it’s the excluded boundary).

Checks for Understanding

(6 minutes — exit ticket, collected)

1. Verify whether solves , showing LHS and RHS.
2. Verify whether satisfies .
3. A solution set is claimed to be for the inequality . Test and to confirm the boundary is open.
4. Find the error: ; working shown as ; .
5. Reasoning. Explain why testing a boundary value is a reliable way to decide whether a circle should be open or closed, without memorising a rule.

Answers: 1. LHS, RHS ✓ yes; 2. LHS, and ✓ yes; 3. : , — excluded (correctly open); : — included; 4. dividing by should reverse the sign, giving , not ; 5. Substituting the exact boundary value into the original inequality shows directly whether that value makes the statement true or false — this is definitive and does not rely on recalling the symbol convention.

Common Misconceptions

Misconception	How to pre-empt it
Substituting into a simplified or intermediate line rather than the original equation/inequality.	Insist verification always uses the original statement, before any solving steps.
Assuming every “find the error” task must contain an error.	Include at least one fully correct piece of working, as in Student A, to keep verification genuine.
Treating boundary testing the same as boundary testing.	Explicitly test the boundary value itself and observe whether the resulting statement is true or false.
Concluding a value “roughly works” without exact arithmetic.	Require exact evaluation of both sides — no estimation during verification.
Believing a failed check means the inequality itself is wrong, rather than the solving.	Separate the two: the inequality is a fixed statement; only the proposed solution set can be wrong.

Enrichment — Competition-Style Problems

E1 (Kangaroo style). Which of , , satisfy ?

Answer

: ✓. : ✓. : ✗. So and only.

E2 (AMC Junior style). For what value of does lie exactly on the boundary of ?

Answer

.

E3 (Challenge). A student claims is the only solution to . Verify this claim and correct it if needed.

Answer

✓ true, but : ✓ also true. The claim is false — the true solution set is , infinitely many values, not just .

E4 (Challenge). Find all integer values of that satisfy both and , by testing candidate integers.

Answer

First: . Second: . Combined: . Integers: .

Homework

1. Verify whether each value is the solution, showing LHS and RHS: (a) for (b) for .
2. Verify whether each value satisfies the inequality: (a) for (b) for .
3. A claimed solution set is for . Test and to confirm whether the boundary is correct.
4. Find the error and give the correct solution: ; ; .
5. Reasoning. A classmate says “if a check fails, the equation must have no solution.” Explain why this reasoning is flawed, using the idea of re-solving versus re-checking.
6. Challenge. Find all integer values of satisfying both and .

Answers: Q1 — (a) LHS, RHS ✓ yes (b) LHS, RHS ✓ yes. Q2 — (a) ✓ yes (b) ✓ yes. Q3 — : ✓ included; : false, correctly excluded — boundary correct. Q4 — dividing by should reverse: , not . Q5 — a failed check means the proposed solution is wrong, not that the equation is unsolvable; the correct response is to re-solve carefully, not to conclude “no solution.” Q6 — ; ; combined ; integers .