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 = 4 x + 8.
Step 2: 4 x + 8 = 24.
Step 3: subtract 8 → 4 x = 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.