O-Level Equations and inequalities: worked solution

Question

Find all solutions of the pair of equations y − x = 4 and 4x² − 6xy − 6y = 4.

y − x = 4 ... (1) 4x² − 6xy − 6y = 4 ... (2) Step 1 - use the LINEAR equation to make one letter the subject. With one linear and one non-linear equation you always substitute from the linear one, because rearranging it produces no squares or fractions. From (1): y = x + 4 Step 2 - substitute y = x + 4 into (2), replacing EVERY y. 4x² − 6x(x + 4) − 6(x + 4) = 4 Step 3 - expand each bracket carefully, watching the minus signs. −6x(x + 4) = −6x² − 24x −6(x + 4) = −6x − 24 So 4x² − 6x² − 24x − 6x − 24 = 4 Step 4 - collect like terms. (4x² − 6x²) + (−24x − 6x) − 24 = 4 −2x² − 30x − 24 = 4 Step 5 - bring everything to one side to get a quadratic equal to zero. −2x² − 30x − 24 − 4 = 0 −2x² − 30x − 28 = 0 Divide every term by −2 to make the x² coefficient positive and the numbers small (dividing by a negative flips all the signs): x² + 15x + 14 = 0 Step 6 - factorise. Look for two numbers that multiply to +14 and add to +15: these are +14 and +1. (x + 14)(x + 1) = 0 So x + 14 = 0 or x + 1 = 0, giving x = −14 or x = −1. Step 7 - find the matching y for EACH x, using the linear equation y = x + 4. If x = −14: y = −14 + 4 = −10 If x = −1: y = −1 + 4 = 3 Pair the values up; do not mix them. Step 8 - check in (2). x = −14, y = −10: 4(196) − 6(−14)(−10) − 6(−10) = 784 − 840 + 60 = 4 ✓ x = −1, y = 3: 4(1) − 6(−1)(3) − 6(3) = 4 + 18 − 18 = 4 ✓