Mathematics • Year 10 • Unit 2 • Lesson 9

Fraction Equations in the Real World

Use LCDs and cross-multiplication on real-world setups — sharing pizza, scaling maps, fuel-economy rates and average speed. Then defend why a restriction matters when the unknown appears in a denominator.

Apply · Real-World Maths

1. Word problems

For each problem: define the variable, set up the fraction equation, solve it, and state the answer in context.

1.1 — Sharing a pizza. A group of friends shares a pizza equally. Mai eats one quarter of the pizza, Sam eats one third, and 5 slices are left over. The whole pizza has x slices.

(a) Write expressions for the number of slices Mai and Sam each ate.
(b) Set up an equation that says Mai's slices + Sam's slices + 5 = total slices.
(c) Clear fractions using the LCD and solve for x.    4 marks

Stuck? x/4 + x/3 + 5 = x. Multiply every term by 12.

1.2 — Map scale. On a hiking map, 2 cm represents 5 km of real distance. Two campsites are x cm apart on the map and 14 km apart in real life.

(a) Write a proportion: x/14 = 2/5.
(b) Cross-multiply and solve for x. State the restriction (if any).    3 marks

Stuck? Cross-multiplying gives 5x = 28.

1.3 — Fuel economy. A car uses 1 litre of fuel for every x km it drives. On a long trip the car uses 24 L to drive 360 km.

(a) Write the equation 1/x = 24/360. State the restriction.
(b) Cross-multiply and solve for x. Interpret the answer (km per litre).    3 marks

1.4 — Average speed. Mai cycles to school at 18 km/h and walks home at 6 km/h. The total distance each way is d km. Her total travel time is 1 hour.

(a) Write expressions for the time taken to cycle and to walk (use time = distance/speed).
(b) Set up the equation total time = 1, then solve for d.    4 marks

Stuck? d/18 + d/6 = 1. LCD = 18.

1.5 — Concentration of juice. A bottle contains 5 mL of cordial in x mL of water. The recipe says 1 mL cordial per 4 mL water. Find x so the bottle matches the recipe.

(a) Write the proportion 5/x = 1/4. State the restriction.
(b) Solve and state x in mL.    3 marks

2. Explain your thinking

Communication, not just numbers. 4 marks

2.1 A classmate solves 6/(x − 3) = 2 and gets the answer x = 3. They are very pleased because the numbers feel clean. Using the lesson rules for restrictions and algebraic fractions, explain (i) why x = 3 cannot be a valid solution, (ii) what the actual restriction is for this equation, and (iii) how to find the real solution. Show the corrected working.

Stuck? If x = 3, then x − 3 = 0 and 6/0 is undefined.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Pizza

(a) Mai = x/4, Sam = x/3. (b) x/4 + x/3 + 5 = x. (c) Multiply every term by 12: 3x + 4x + 60 = 12x → 7x + 60 = 12x → 60 = 5x → x = 12 slices. Check: 12/4 + 12/3 + 5 = 3 + 4 + 5 = 12 ✓.

1.2 — Map scale

(a) x/14 = 2/5 (no restriction needed since both denominators are constants). (b) Cross-multiply: 5x = 28 → x = 5.6 cm.

1.3 — Fuel economy

(a) Restriction: x ≠ 0. (b) Cross-multiply: 360 = 24x → x = 15. The car gets 15 km per litre.

1.4 — Average speed

(a) Cycle time = d/18 h; walk time = d/6 h. (b) d/18 + d/6 = 1. LCD = 18: d + 3d = 18 → 4d = 18 → d = 4.5 km. Check: 4.5/18 + 4.5/6 = 0.25 + 0.75 = 1 h ✓.

1.5 — Cordial

(a) 5/x = 1/4 with x ≠ 0. (b) Cross-multiply: 20 = x → x = 20 mL of water.

2.1 — Explain (sample response)

(i) When x = 3, the denominator x − 3 becomes 0, and division by 0 is undefined — so the original equation has no meaning at x = 3. It cannot be a valid solution no matter what the rest of the algebra gives. (ii) The restriction for this equation is x ≠ 3. (iii) Correctly solve by multiplying both sides by (x − 3): 6 = 2(x − 3) → 6 = 2x − 6 → 12 = 2x → x = 6. Check x ≠ 3 ✓. Verify: 6/(6 − 3) = 6/3 = 2 ✓.

Marking: 1 mark for explaining division by zero; 1 mark for the restriction x ≠ 3; 1 mark for correct algebra to x = 6; 1 mark for the substitution check.