Mathematics • Year 10 • Unit 2 • Lesson 9

Equations with Fractions — Skill Drill

Build fluency with the three methods from Lesson 9: clearing denominators using the LCD, cross-multiplication, and handling algebraic fractions with restrictions (x ≠ value that makes the denominator zero).

Build · I Do / We Do / You Do

1. I do — fully worked example

Clearing fractions by multiplying every term by the LCD. Read each reason.

Problem. Solve x/3 + x/6 = 5.

Step 1 — Find the LCD of the denominators 3 and 6.

LCD = 6

Reason: 6 is the smallest number both 3 and 6 divide into.

Step 2 — Multiply every term by 6.

6 × (x/3) + 6 × (x/6) = 6 × 5

2x + x = 30

Reason: 6/3 = 2 and 6/6 = 1, so the denominators are gone. EVERY term gets multiplied, including the 5.

Step 3 — Solve the simpler linear equation.

3x = 30   →   x = 10

Step 4 — Check by substitution.

10/3 + 10/6 = 20/6 + 10/6 = 30/6 = 5 ✓

Answer: x = 10.

Stuck? Revisit lesson § "Clearing Fractions with the LCD" — Worked Example 1.

2. We do — fill in the missing steps

Solving by cross-multiplication. Fill in each blank. 5 marks

Problem. Solve 4/x = 2/7. State any restrictions.

Step 1 — Restriction: x ≠ _____ (the value that would make a denominator zero).

Step 2 — Cross-multiply:

4 × _____ = 2 × _____

Step 3 — Simplify both sides:

_____ = _____ x

Step 4 — Divide both sides by 2:

x = _____

Step 5 — Check restriction: Is your x equal to 0? _____ → valid solution.

Stuck? Revisit lesson § "Cross-Multiplication" — Worked Example 2.

3. You do — independent practice

Show every step. State restrictions when there is a variable in the denominator.

Foundation — single skill

3.1 Solve x/5 = 4.    1 mark

3.2 Solve x/4 + 1 = 3.    1 mark

3.3 Find the LCD of x/2, x/3 and x/4.    1 mark

3.4 Solve 3/x = 2/5 by cross-multiplying. State the restriction.    1 mark

Standard — combine tools

3.5 Solve x/2 + x/3 = 10 by clearing fractions with the LCD.    2 marks

3.6 Solve (x + 1)/2 = (x − 1)/3 by clearing fractions.    2 marks

Extension — algebraic fractions and restrictions

3.7 Solve 5/(x − 2) = 3. State the restriction.    3 marks

3.8 Solve x/2 + 4 = x/3 + 6. (Choose method: LCD = 6 works well.)    3 marks

Stuck on 3.7? Multiply both sides by (x − 2), then expand and solve.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (4/x = 2/7)

Step 1: x ≠ 0. Step 2: 4 × 7 = 2 × x. Step 3: 28 = 2x. Step 4: x = 14. Step 5: 14 ≠ 0, so the solution is valid. Check: 4/14 = 2/7 ✓.

3.1 — x/5 = 4

Multiply by 5: x = 20.

3.2 — x/4 + 1 = 3

Subtract 1: x/4 = 2. Multiply by 4: x = 8.

3.3 — LCD of 2, 3, 4

LCD = 12 (smallest multiple of 2, 3 and 4).

3.4 — 3/x = 2/5

Restriction: x ≠ 0. Cross-multiply: 3 × 5 = 2 × x → 15 = 2x → x = 7.5.

3.5 — x/2 + x/3 = 10

LCD = 6. Multiply every term by 6: 3x + 2x = 60 → 5x = 60 → x = 12. Check: 12/2 + 12/3 = 6 + 4 = 10 ✓.

3.6 — (x + 1)/2 = (x − 1)/3

LCD = 6. Multiply every term: 3(x + 1) = 2(x − 1). Expand: 3x + 3 = 2x − 2. Subtract 2x: x + 3 = −2. Subtract 3: x = −5. Check: (−5 + 1)/2 = −2 and (−5 − 1)/3 = −2 ✓.

3.7 — 5/(x − 2) = 3

Restriction: x ≠ 2. Multiply both sides by (x − 2): 5 = 3(x − 2). Expand: 5 = 3x − 6. Add 6: 11 = 3x → x = 11/3 ≈ 3.67. Valid (≠ 2). Check: 5/(11/3 − 2) = 5/(5/3) = 3 ✓.

3.8 — x/2 + 4 = x/3 + 6

LCD = 6. Multiply every term: 3x + 24 = 2x + 36. Subtract 2x: x + 24 = 36. Subtract 24: x = 12. Check: 12/2 + 4 = 10 and 12/3 + 6 = 10 ✓.

HSCScience • Mathematics • Year 10 • Unit 2 • Lesson 9 • Worksheet 1 (Build)