Mathematics • Year 10 • Unit 2 • Lesson 9
Equations with Fractions — Mixed Challenge
Mix the three methods (LCD, cross-multiplication, algebraic fractions with restrictions) and decide which one fits each problem fastest. Spot a classmate's missing-restriction error, then design a fraction equation with a chosen solution.
1. Mixed problems — choose the right tool
Decide whether to use LCD, cross-multiplication or algebraic-fraction methods before writing. State any restrictions and check. 3 marks each
1.1 Solve x/4 + x/6 = 5.
1.2 Solve 2/x = 5/8. State the restriction.
1.3 Solve (2x + 1)/3 = (x + 4)/2. (LCD = 6.)
1.4 Solve 7/(x + 1) = 2. State the restriction.
1.5 Solve x/3 − x/5 = 4. (LCD = 15.)
1.6 Solve (x − 2)/4 + (x + 1)/3 = 5. (LCD = 12, expand carefully.)
2. Find the mistake
A Year 10 student solved 3/(x − 4) = 1 and stopped at x = 7 without checking the restriction. Their working is below. One line contains an error or omission. Fix it. 3 marks
Student's working:
Line 1: 3/(x − 4) = 1
Line 2: 3 = 1 × (x − 4)
Line 3: 3 = x − 4
Line 4: x = 7
(a) Which step is missing or incorrect?
(b) Explain why it is needed.
(c) Show the corrected working including the restriction and a substitution check.
Stuck? The arithmetic is right but the student never wrote x ≠ 4.3. Open-ended challenge — design your own
Many valid answers. 4 marks
3.1 Design a fraction equation that uses (i) at least one variable in a denominator, (ii) two different denominators that need an LCD step, AND (iii) a non-zero restriction you must state. The equation must have x = 6 as its solution.
Submit:
(a) Your equation, written cleanly.
(b) The restriction.
(c) Full working that solves to x = 6.
(d) A substitution check that the equation holds.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — x/4 + x/6 = 5
LCD = 12. 3x + 2x = 60 → 5x = 60 → x = 12. Check: 12/4 + 12/6 = 3 + 2 = 5 ✓.
1.2 — 2/x = 5/8
Restriction: x ≠ 0. Cross-multiply: 16 = 5x → x = 16/5 = 3.2.
1.3 — (2x + 1)/3 = (x + 4)/2
LCD = 6. Multiply every term by 6: 2(2x + 1) = 3(x + 4) → 4x + 2 = 3x + 12 → x = 10. Check: (21)/3 = 7 and (14)/2 = 7 ✓.
1.4 — 7/(x + 1) = 2
Restriction: x ≠ −1. Multiply both sides by (x + 1): 7 = 2(x + 1) → 7 = 2x + 2 → 5 = 2x → x = 2.5. ≠ −1 ✓. Check: 7/(2.5 + 1) = 7/3.5 = 2 ✓.
1.5 — x/3 − x/5 = 4
LCD = 15. Multiply: 5x − 3x = 60 → 2x = 60 → x = 30. Check: 30/3 − 30/5 = 10 − 6 = 4 ✓.
1.6 — (x − 2)/4 + (x + 1)/3 = 5
LCD = 12. Multiply every term: 3(x − 2) + 4(x + 1) = 60 → 3x − 6 + 4x + 4 = 60 → 7x − 2 = 60 → 7x = 62 → x = 62/7 ≈ 8.86. Check: (62/7 − 2)/4 + (62/7 + 1)/3 = (48/7)/4 + (69/7)/3 = 12/7 + 23/7 = 35/7 = 5 ✓.
2 — Find the mistake
(a) Missing restriction at the very start.
(b) Because x appears in a denominator, the lesson rule requires us to state the value that would make the denominator zero before solving. Here x − 4 = 0 when x = 4, so x ≠ 4 must be written down. If our final answer accidentally lands on x = 4, it would have to be rejected.
(c) Corrected:
3/(x − 4) = 1, restriction: x ≠ 4.
Multiply both sides by (x − 4): 3 = x − 4 → x = 7 ≠ 4 ✓. Check: 3/(7 − 4) = 3/3 = 1 ✓.
3 — Open-ended (sample solution)
Sample equation: 12/x + 1/3 = 7/3.
Restriction: x ≠ 0.
Working: LCD = 3x. Multiply every term: 36 + x = 7x → 36 = 6x → x = 6 ✓.
Check: 12/6 + 1/3 = 2 + 1/3 = 7/3 ✓.
Marking: 1 for a valid equation meeting all three criteria; 1 for the correct restriction; 1 for working that lands on x = 6; 1 for a clean substitution check.