Mathematics • Year 8 • Unit 2 • Lesson 18

Simultaneous Equations — Substitution Method

Master the four-step substitution method: Isolate → Substitute → Solve → Back-substitute. One worked example with a pre-isolated variable, one guided example with blanks where you isolate first, then eight independent problems graduated from "y already isolated" up to "isolate then expand a bracket".

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every line. The substitution method swaps one variable for an equal expression, leaving one equation with one unknown.

Problem. Solve y = x + 3 and 2x + y = 15 by substitution.

Step 1 — Isolate.

Eq 1 is already isolated: y = x + 3.

Reason: when one variable is already alone on a side, there is no rearranging to do.

Step 2 — Substitute (x + 3) for y in the OTHER equation. Use brackets!

2x + (x + 3) = 15

Reason: replacing equals with equals keeps the equation true. Brackets stop sign errors.

Step 3 — Solve the single-variable equation.

3x + 3 = 15 → 3x = 12 → x = 4

Step 4 — Back-substitute x = 4 into the isolated equation to find y.

y = 4 + 3 = 7

Step 5 — VERIFY in BOTH original equations.

Eq 1: 7 = 4 + 3 ✓. Eq 2: 2(4) + 7 = 15 ✓.

Answer: (x, y) = (4, 7).

Stuck? Revisit lesson § "The Big Idea — Substitution". Steps: Isolate → Substitute → Solve → Back-substitute → Verify.

2. We do — fill in the missing steps

Same shape, with blanks. Here you have to ISOLATE first because no variable is alone yet. 5 marks

Problem. Solve x + y = 7 and 2x − y = 5 by substitution.

Step 1 — Isolate y in Eq 1 (its coefficient is 1, so this is the easiest).

x + y = 7 → y = ______ − ______

Step 2 — Substitute that expression for y in Eq 2. Use brackets!

2x − (______ − ______) = 5

Step 3 — Expand the brackets carefully — the minus sign distributes:

2x − 7 + ______ = 5 → ______x = ______ → x = ______

Step 4 — Back-substitute to find y:

y = 7 − ______ = ______

Step 5 — Verify: Eq 1: ______ + ______ = ______ ✓ ? Eq 2: 2(______) − ______ = ______ ✓ ?

Answer: (x, y) = (______, ______).

Stuck? Watch the minus sign in step 3: −(7 − x) = −7 + x, NOT −7 − x.

3. You do — independent practice

Show every step. 3.1–3.3 are foundation (one variable already isolated). 3.4–3.6 are standard (isolate first, no brackets). 3.7–3.8 are extension (isolate then handle a tricky substitution).

Foundation — one variable already isolated

3.1 y = 2x and x + y = 12. Solve. 2 marks

3.2 y = x + 1 and 3x + y = 17. Solve. 2 marks

3.3 x = 2y and x + 3y = 25. Solve. 2 marks

Standard — isolate first (no brackets)

3.4 x + y = 10 and 2x + y = 16. Isolate y in Eq 1, then substitute. 3 marks

3.5 2x + y = 11 and 3x − y = 4. Isolate y in Eq 1, then substitute. 3 marks

3.6 x + 2y = 13 and 3x − y = 4. Choose the easiest variable to isolate, then solve. 3 marks

Extension — minus-sign / brackets care

3.7 x − y = 2 and 3x + y = 14. Isolate y in Eq 1, then substitute into Eq 2. Be careful with signs. 3 marks

3.8 2x − y = 1 and 4x + 3y = 22. Isolate y in Eq 1 (y = 2x − 1), substitute, expand carefully, then solve. 3 marks

Stuck on 3.8? After substituting: 4x + 3(2x − 1) = 22. Expand FIRST: 4x + 6x − 3 = 22.

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Answers — Do not peek before attempting

Section 2 — We do x + y = 7 and 2x − y = 5

Step 1: y = 7 − x.
Step 2: 2x − (7 − x) = 5.
Step 3: 2x − 7 + x = 5 → 3x = 12 → x = 4.
Step 4: y = 7 − 4 = 3.
Step 5: Eq 1: 4 + 3 = 7 ✓. Eq 2: 2(4) − 3 = 5 ✓.
Answer: (x, y) = (4, 3).

3.1 — y = 2x, x + y = 12

Sub: x + 2x = 12 → 3x = 12 → x = 4. Back-sub: y = 2(4) = 8. Solution: (4, 8). Check: 4 + 8 = 12 ✓.

3.2 — y = x + 1, 3x + y = 17

Sub: 3x + (x + 1) = 17 → 4x + 1 = 17 → 4x = 16 → x = 4. y = 4 + 1 = 5. Solution: (4, 5). Check: 3(4) + 5 = 17 ✓.

3.3 — x = 2y, x + 3y = 25

Sub: 2y + 3y = 25 → 5y = 25 → y = 5. x = 2(5) = 10. Solution: (10, 5). Check: 10 + 3(5) = 25 ✓.

3.4 — x + y = 10, 2x + y = 16

Isolate y in Eq 1: y = 10 − x. Sub: 2x + (10 − x) = 16 → x + 10 = 16 → x = 6. y = 10 − 6 = 4. Solution: (6, 4). Check: 2(6) + 4 = 16 ✓.

3.5 — 2x + y = 11, 3x − y = 4

Isolate y in Eq 1: y = 11 − 2x. Sub: 3x − (11 − 2x) = 4 → 3x − 11 + 2x = 4 → 5x = 15 → x = 3. y = 11 − 2(3) = 5. Solution: (3, 5). Check: 3(3) − 5 = 4 ✓.

3.6 — x + 2y = 13, 3x − y = 4

Easiest: isolate y in Eq 2 (coefficient −1): y = 3x − 4. Sub into Eq 1: x + 2(3x − 4) = 13 → x + 6x − 8 = 13 → 7x = 21 → x = 3. y = 3(3) − 4 = 5. Solution: (3, 5). Check: 3 + 2(5) = 13 ✓.

3.7 — x − y = 2, 3x + y = 14

Isolate y in Eq 1: y = x − 2. Sub: 3x + (x − 2) = 14 → 4x − 2 = 14 → 4x = 16 → x = 4. y = 4 − 2 = 2. Solution: (4, 2). Check: 4 − 2 = 2 ✓.

3.8 — 2x − y = 1, 4x + 3y = 22

Isolate y in Eq 1: y = 2x − 1. Sub: 4x + 3(2x − 1) = 22 → 4x + 6x − 3 = 22 → 10x = 25 → x = 5/2 = 2.5. y = 2(2.5) − 1 = 4. Solution: (2.5, 4). Check: 4(2.5) + 3(4) = 10 + 12 = 22 ✓.