Elimination method · 4 marks

The method used in solving this problem is called the Elimination Method because we eliminate one of the variables to obtain a linear equation in the other variable. In the example above, we eliminated y. Try the same problem by eliminating x instead of y. Also try to solve the two equations using the Substitution Method. Assess the pros and cons of each method.

1. Try the same problem by eliminating x instead of y.
2. Also try to solve the two equations using the Substitution Method.
3. Assess the pros and cons of each method.

Answer: Eliminating x (multiply by 7 and 9) and substitution both give x = 2000, y = 4000, so the incomes are ₹18,000 and ₹14,000 again. Elimination avoids fractions here; substitution is best when a variable has coefficient 1.

Step-by-step solution

Given: Incomes 9x, 7x; expenditures 4y, 3y; 9x − 4y = 2000 … (1); 7x − 3y = 2000 … (2)

Idea: The equations are 9x − 4y = 2000 … (1) and 7x − 3y = 2000 … (2) (incomes 9x, 7x; expenditures 4y, 3y). Any correct method must give the same answer.

1. Try the same problem by eliminating x instead of y.

Make the x-coefficients equal (63): (1) × 7: 63x − 28y = 14000 … (3); (2) × 9: 63x − 27y = 18000 … (4).½ mark
(4) − (3): y = 4000. Then (1): 9x = 2000 + 16000 = 18000 ⇒ x = 2000.1 mark

x = 2000, y = 4000

2. Also try to solve the two equations using the Substitution Method.

From (2): 7x = 2000 + 3y ⇒ x = 2000 + 3y7.½ mark
Into (1): 9(2000 + 3y7) − 4y = 2000. Multiply by 7: 18000 + 27y − 28y = 14000 ⇒ y = 4000; then x = 2000 + 120007 = 2000.1 mark

x = 2000, y = 4000

3. Assess the pros and cons of each method.

Elimination: pro: no fractions here, just whole-number multiples, and it is quick when coefficients can be matched easily. Con: you may need large multipliers (×7 and ×9 gave 63) and must keep careful track of signs when subtracting.½ mark
Substitution: pro: very quick when one variable already has coefficient 1 (like y = 4x or x + 2y = 3). Con: here no coefficient is 1, so it brings in the fraction 2000 + 3y7 and messier algebra. Both methods give the same answer; choose the one that keeps the numbers simple.½ mark

Same answer; choose the method that keeps the working simplest.

Both ways give x = 2000, y = 4000 (incomes ₹18,000 and ₹14,000). Elimination suits this pair better; substitution is better when some variable has coefficient 1.

Check: 9(2000) − 4(4000) = 18000 − 16000 = 2000 ✓; 7(2000) − 3(4000) = 14000 − 12000 = 2000 ✓.

Answer to write in the exam

(1) × 7: 63x − 28y = 14000 … (3)

(2) × 9: 63x − 27y = 18000 … (4)

(4) − (3): y = 4000

(1): 9x − 16000 = 2000 ⇒ x = 2000

∴ Incomes: 9 × 2000 = ₹18,000 and 7 × 2000 = ₹14,000

From (2): x = 2000 + 3y7

In (1): 9(2000 + 3y)7 − 4y = 2000

18000 + 27y − 28y = 14000 ⇒ y = 4000

x = 140007 = 2000

Elimination: no fractions here; but may need large multipliers and careful signs

Substitution: best when a coefficient is 1; here it gives fractions

∴ Both give x = 2000, y = 4000

Common mistakes that cost marks

Multiplying only the left side of an equation by 7 or 9 and forgetting the right side.
In substitution, forgetting to multiply 4y by 7 when clearing the denominator.
Thinking different methods can give different answers. A pair with a unique solution has only one answer.

How this can come in the exam

Short answer (3 marks)

Solve 4x + 3y = 25, 3x + 4y = 24 by eliminating x.

Show answer

×3 and ×4: 12x + 9y = 75, 12x + 16y = 96 (1 mark). Subtract: 7y = 21 ⇒ y = 3 (1 mark). 4x = 16 ⇒ x = 4 (1 mark).

Try one yourself

Solve 5x − 2y = 11, 3x + 4y = 4 by elimination.