Mathematics • Year 7 • Unit 2 • Lesson 15
Equations with Brackets
Build the basics: expand brackets using the distributive law a(b + c) = ab + ac, then solve the resulting two-step equation. Know when to divide first instead.
1. I do — fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. Solve 2(3x + 1) = 20.
Step 1 — Expand the brackets (distributive law).
2(3x + 1) = 2 × 3x + 2 × 1 = 6x + 2
Reason: the 2 outside multiplies BOTH terms inside, not just the first one. a(b + c) = ab + ac.
Step 2 — Rewrite the equation in standard two-step form.
6x + 2 = 20
Reason: now it's a familiar ax + b = c equation — use SADMEP from Lesson 14.
Step 3 — Undo the +2 first (subtract 2 from BOTH sides).
6x + 2 − 2 = 20 − 2 → 6x = 18
Step 4 — Undo the ×6 (divide BOTH sides by 6).
6x ÷ 6 = 18 ÷ 6 → x = 3
Step 5 — Check by substitution INTO the ORIGINAL.
2(3 × 3 + 1) = 2(9 + 1) = 2(10) = 20 ✓
Answer: x = 3.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. Solve 4(x + 2) = 24.
Step 1 — Expand:
4(x + 2) = 4 × ____ + 4 × ____ = ____ x + ____
Step 2 — Rewrite the equation:
____ x + ____ = 24
Step 3 — Subtract ____ from both sides:
____ x = ____
Step 4 — Divide both sides by ____ :
x = ____
Step 5 — Check in the ORIGINAL:
4(____ + 2) = 4(____) = ____ ✓ matches RHS
3. You do — independent practice
Show your working — at minimum the expand step and the SADMEP steps. The first four are foundation, the middle two are standard, and the last two are extension.
Foundation — single brackets, easy numbers
3.1 Expand 3(x + 5). (Don't solve — just expand.) 1 mark
3.2 Solve 2(x + 5) = 14. 1 mark
3.3 Solve 2(x − 5) = 12. 1 mark
3.4 Solve 5(x + 3) = 35 — once by EXPANDING first, and once by DIVIDING by 5 first. Confirm both routes give the same x. 1 mark
Standard — coefficient inside the brackets
3.5 Solve 3(2x + 1) = 27. 2 marks
3.6 Solve 4(2x − 3) = 20. 2 marks
Extension — brackets on TOP of a fraction
3.7 Solve (x + 3)⁄4 = 5. (Hint: multiply BOTH sides by 4 first to get rid of the denominator.) 2 marks
3.8 Solve 2(x + 4) = 18. Solve it BOTH ways: (i) divide by 2 first, (ii) expand first. Which way feels quicker, and why? 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (4(x + 2) = 24)
Step 1: 4(x + 2) = 4 × x + 4 × 2 = 4x + 8.
Step 2: 4x + 8 = 24.
Step 3: subtract 8 → 4x = 16.
Step 4: divide both sides by 4 → x = 4.
Step 5: 4(4 + 2) = 4(6) = 24 ✓.
3.1 — Expand 3(x + 5)
3(x + 5) = 3 × x + 3 × 5 = 3x + 15.
3.2 — 2(x + 5) = 14
Expand: 2x + 10 = 14. Subtract 10: 2x = 4. Divide by 2: x = 2. Check: 2(2 + 5) = 2(7) = 14 ✓.
3.3 — 2(x − 5) = 12
Expand: 2x − 10 = 12. Add 10: 2x = 22. Divide by 2: x = 11. Check: 2(11 − 5) = 2(6) = 12 ✓.
3.4 — 5(x + 3) = 35 (two routes)
EXPAND first: 5x + 15 = 35 → 5x = 20 → x = 4.
DIVIDE by 5 first: x + 3 = 7 → x = 4.
Both routes agree. Check: 5(4 + 3) = 5(7) = 35 ✓.
3.5 — 3(2x + 1) = 27
Expand: 6x + 3 = 27. Subtract 3: 6x = 24. Divide by 6: x = 4. Check: 3(2(4) + 1) = 3(9) = 27 ✓. (Or divide by 3 first: 2x + 1 = 9 → 2x = 8 → x = 4.)
3.6 — 4(2x − 3) = 20
Expand: 8x − 12 = 20. Add 12: 8x = 32. Divide by 8: x = 4. Check: 4(2(4) − 3) = 4(5) = 20 ✓. (Or divide by 4 first: 2x − 3 = 5 → 2x = 8 → x = 4.)
3.7 — (x + 3)⁄4 = 5
Multiply both sides by 4: x + 3 = 20. Subtract 3: x = 17. Check: (17 + 3)⁄4 = 20⁄4 = 5 ✓.
3.8 — 2(x + 4) = 18 (two ways)
(i) Divide by 2: x + 4 = 9 → x = 5.
(ii) Expand: 2x + 8 = 18 → 2x = 10 → x = 5.
Dividing by 2 first is usually quicker here, because 18 ÷ 2 = 9 cleanly and we avoid two larger numbers (2x + 8 and 10) on the way. Choose "divide first" when the number outside divides into the RHS without a fraction.