Solve the following pair of equations by substitution method:
7x − 15y = 2 … (1)
x + 2y = 3 … (2)
Answer: x = 4929, y = 1929.
Step-by-step solution
Idea: Make one variable the subject of the simpler equation (x = 3 − 2y from (2)), put that into the other equation so only y is left, solve for y, then go back for x.
- Step 1: From (2), x = 3 − 2y … (3).½ mark
- Step 2: Every common solution must satisfy (3), so substitute it in (1): 7(3 − 2y) − 15y = 2 ⇒ 21 − 14y − 15y = 2 ⇒ −29y = −19 ⇒ y = 1929.1 mark
- Step 3: Put y in (3): x = 3 − 2(1929) = 8729 − 3829 = 4929.1 mark
- Verify (the “try it for yourself” check): (1) 7 × 4929 − 15 × 1929 = 343 − 28529 = 5829 = 2 ✓; (2) 4929 + 3829 = 8729 = 3 ✓.½ mark
x = 49/29, y = 19/29.
Check: Both equations are satisfied exactly (see the last step), so the fractions are correct even though they look unusual.
Answer to write in the exam
From (2): x = 3 − 2y … (3)
Put in (1): 7(3 − 2y) − 15y = 2
21 − 29y = 2 ⇒ −29y = −19 ⇒ y = 1929
From (3): x = 3 − 3829 = 4929
∴ x = 4929, y = 1929
Common mistakes that cost marks
- Expanding 7(3 − 2y) as 21 − 2y; the 7 multiplies both terms: 21 − 14y.
- Sign error: −14y − 15y = −29y, not −y.
- Writing 3 − 3829 = 3529 instead of changing 3 to 8729 first.
How this can come in the exam
MCQ (1 mark)
Solving x + 3y = 7 and 2x − y = 0 by substitution gives
- x = 1, y = 2
- x = 2, y = 1
- x = 1, y = −2
- x = 4, y = 1
Show answer
(A) x = 1, y = 2
y = 2x ⇒ x + 6x = 7 ⇒ x = 1, y = 2.
Short answer (3 marks)
Solve by substitution: 3x − 5y = 4, x − 2y = 1.
Show answer
x = 1 + 2y (½). 3(1 + 2y) − 5y = 4 ⇒ 3 + y = 4 ⇒ y = 1 (1½). x = 3 (½). Check: 9 − 5 = 4 ✓ (½).Try one yourself
Solve by substitution: 5x − 2y = 1, x + y = 4.
Show answer
y = 4 − x ⇒ 5x − 8 + 2x = 1 ⇒ x = 97, y = 197.
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