Mathematics • Year 10 • Unit 2 • Lesson 8
Solving Linear Equations — Mixed Challenge
Bring together every tool from Lesson 8: balance rule, reverse BIDMAS, brackets-first, variables on both sides. Choose the right tool, spot a classmate's slip with sign-tracking, then design your own equation with a target solution.
1. Mixed problems — choose the right tool
Each question uses a different idea from Lesson 8. Show every step including a check. 3 marks each
1.1 Solve 7 − 2x = 1.
1.2 Solve (x + 3)/4 = 5. (Hint: multiply both sides by 4 first.)
1.3 Solve 2x + 8 = 5x − 4.
1.4 Solve −3(x − 2) = 12. (Watch the negative outside the bracket.)
1.5 Solve 5(2x − 3) = 4(x + 3). Expand both sides, then solve.
1.6 A rectangle has length (3x − 1) cm and width (x + 2) cm. Its perimeter is 38 cm. Find x and state the dimensions.
2. Find the mistake
A Year 10 student is solving 3(x − 4) = 2x + 5. One line contains a mistake. Find it, explain why, then redo correctly. 3 marks
Student's working:
Line 1: 3(x − 4) = 2x + 5
Line 2: 3x − 4 = 2x + 5
Line 3: x − 4 = 5
Line 4: x = 9
(a) Which line is wrong?
(b) Explain the mistake.
(c) Show the corrected working, including the corrected final answer.
Stuck? Revisit lesson § "Brackets First" — "Multiply every term".3. Open-ended challenge — design an equation
Many valid answers. Be creative but show every number. 4 marks
3.1 Design a linear equation that uses all four of: brackets, variables on both sides, a negative coefficient, and a non-integer solution. The equation must have x = 5/2 (that is, x = 2.5) as its only solution.
In your submission, include:
(i) Your equation, written cleanly.
(ii) Step-by-step solving that produces x = 2.5.
(iii) A substitution check that both sides agree.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 7 − 2x = 1
Subtract 7: −2x = −6. Divide by −2: x = 3. Check: 7 − 2(3) = 7 − 6 = 1 ✓.
1.2 — (x + 3)/4 = 5
Multiply both sides by 4: x + 3 = 20. Subtract 3: x = 17. Check: (17 + 3)/4 = 20/4 = 5 ✓.
1.3 — 2x + 8 = 5x − 4
Subtract 2x: 8 = 3x − 4. Add 4: 12 = 3x. Divide by 3: x = 4. Check: 2(4) + 8 = 16 and 5(4) − 4 = 16 ✓.
1.4 — −3(x − 2) = 12
Expand: −3x + 6 = 12. Subtract 6: −3x = 6. Divide by −3: x = −2. Check: −3(−2 − 2) = −3(−4) = 12 ✓.
1.5 — 5(2x − 3) = 4(x + 3)
Expand: 10x − 15 = 4x + 12. Subtract 4x: 6x − 15 = 12. Add 15: 6x = 27. Divide by 6: x = 27/6 = 4.5. Check: 5(2(4.5) − 3) = 5(6) = 30 and 4(4.5 + 3) = 4(7.5) = 30 ✓.
1.6 — Rectangle
2(3x − 1) + 2(x + 2) = 38 → 6x − 2 + 2x + 4 = 38 → 8x + 2 = 38 → 8x = 36 → x = 4.5. Length = 3(4.5) − 1 = 12.5 cm; width = 4.5 + 2 = 6.5 cm. Check: 2(12.5) + 2(6.5) = 25 + 13 = 38 ✓.
2 — Find the mistake
(a) Line 2.
(b) The student only multiplied the x by 3 and forgot to distribute the 3 to the −4. The lesson rule is "multiply every term": 3(x − 4) = 3x − 12, not 3x − 4.
(c) Corrected:
3(x − 4) = 2x + 5
3x − 12 = 2x + 5 (correct expansion)
x − 12 = 5 (subtract 2x)
x = 17 (add 12). Check: 3(17 − 4) = 3(13) = 39 and 2(17) + 5 = 34 + 5 = 39 ✓.
3 — Open-ended (sample solution)
Sample equation: 2(x − 1) = −x + 5.5.
Solve: 2x − 2 = −x + 5.5 → add x → 3x − 2 = 5.5 → add 2 → 3x = 7.5 → divide by 3 → x = 2.5 ✓.
Check: LHS = 2(2.5 − 1) = 2(1.5) = 3. RHS = −2.5 + 5.5 = 3 ✓.
Criteria met: brackets ✓ (left side), variables on both sides ✓ (x left, −x right), negative coefficient ✓ (the −x), non-integer solution ✓ (2.5).
Marking: 1 for an equation meeting all four criteria; 1 for correct step-by-step solving; 1 for the bracket expansion / sign handling shown; 1 for a successful substitution check.