Mathematics • Year 10 • Unit 2 • Lesson 8

Solving Linear Equations — Skill Drill

Build fluency with the three core tools from Lesson 8: the balance rule (same operation on both sides), reverse BIDMAS (undo + and − before × and ÷), and brackets-first when the equation contains parentheses. One worked example, one guided trace, then eight graduated problems.

Build · I Do / We Do / You Do

1. I do — fully worked example

A two-step equation solved using the balance rule and reverse BIDMAS. Read every reason — it tells you why each step is legal, not just what to do.

Problem. Solve 5x − 8 = 27.

Step 1 — Spot the operation chain on x.

x → ×5 → 5x → −8 → 27

Reason: x has been multiplied by 5 then had 8 subtracted. Reverse BIDMAS undoes − first, then ×.

Step 2 — Add 8 to both sides (undo the −8).

5x − 8 + 8 = 27 + 8

5x = 35

Reason: balance rule — same operation on both sides keeps the equation true.

Step 3 — Divide both sides by 5 (undo the ×5).

5x ÷ 5 = 35 ÷ 5

x = 7

Step 4 — Check by substitution.

5(7) − 8 = 35 − 8 = 27 ✓

Answer: x = 7.

Stuck? Revisit lesson § "Reverse BIDMAS" — Worked Example 2.

2. We do — fill in the missing steps

Same structure as Section 1, with the working faded. Fill in each blank. 5 marks

Problem. Solve 3(2x − 4) = 18.

Step 1 — Expand the brackets (distribute the 3):

3 × 2x = _____ 3 × (−4) = _____

Expanded equation: _____ x − _____ = 18

Step 2 — Add _____ to both sides:

6x = _____

Step 3 — Divide both sides by _____:

x = _____

Step 4 — Check by substitution: 3(2 × _____ − 4) = 3( _____ ) = _____ ✓

Stuck? Revisit lesson § "Brackets First" — Worked Example 3, Step 1.

3. You do — independent practice

Show every step. Foundation = one-step. Standard = two-step or brackets. Extension = variables on both sides.

Foundation — single skill

3.1 Solve x + 9 = 16. 1 mark

3.2 Solve x − 7 = 4. 1 mark

3.3 Solve 4x = 32. 1 mark

3.4 Solve x ÷ 6 = 5 (that is, x/6 = 5). 1 mark

Standard — two-step / brackets

3.5 Solve 2x + 5 = 17. Show the inverse operations. 2 marks

3.6 Solve 4(x − 2) = 20. Show the expansion. 2 marks

Extension — variables on both sides

3.7 Solve 3x + 7 = x + 15. Subtract the smaller variable term first. 3 marks

3.8 Solve 3(2x − 1) = 2(x + 5). Expand both sides, then solve. 3 marks

Stuck on 3.7 or 3.8? Revisit lesson § "Variables to One Side" — subtract the smaller coefficient to avoid negatives.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (3(2x − 4) = 18)

Step 1: 3 × 2x = 6x; 3 × (−4) = −12; expanded equation: 6x − 12 = 18.
Step 2: Add 12 to both sides → 6x = 30.
Step 3: Divide both sides by 6 → x = 5.
Step 4: Check: 3(2 × 5 − 4) = 3( 6 ) = 18 ✓.

3.1 — x + 9 = 16

Subtract 9: x = 7. Check: 7 + 9 = 16 ✓.

3.2 — x − 7 = 4

Add 7: x = 11. Check: 11 − 7 = 4 ✓.

3.3 — 4x = 32

Divide by 4: x = 8. Check: 4(8) = 32 ✓.

3.4 — x/6 = 5

Multiply both sides by 6: x = 30. Check: 30/6 = 5 ✓.

3.5 — 2x + 5 = 17

Subtract 5: 2x = 12. Divide by 2: x = 6. Check: 2(6) + 5 = 17 ✓.

3.6 — 4(x − 2) = 20

Expand: 4x − 8 = 20. Add 8: 4x = 28. Divide by 4: x = 7. Check: 4(7 − 2) = 4(5) = 20 ✓.

3.7 — 3x + 7 = x + 15

Subtract x: 2x + 7 = 15. Subtract 7: 2x = 8. Divide by 2: x = 4. Check: 3(4) + 7 = 19 and 4 + 15 = 19 ✓.

3.8 — 3(2x − 1) = 2(x + 5)

Expand: 6x − 3 = 2x + 10. Subtract 2x: 4x − 3 = 10. Add 3: 4x = 13. Divide by 4: x = 13/4 = 3.25. Check: 3(2(3.25) − 1) = 3(5.5) = 16.5 and 2(3.25 + 5) = 2(8.25) = 16.5 ✓.