If, 3x + y = 14, and 2xy = 32, and 3x > y, then find

Solution

2xy = 32 So, xy = (32/2) = 16 ......(i) 3x + y = 14 On squaring the above equation, we get, (3x + y)² = 14² Or, 9x² + y² + 2 × 3x × y = 196 Or, 9x² + y² + 6 × 16 = 196 (since, xy = 16) Or, 9x² + y² = 196 - 96 So, 9x² + y² = 100 ....(ii) (3x - y)² = 9x² + y² - 2 × 3x × y = 100 - 6 × 16 = 100 - 96 = 4 {using values from equations (i) and (ii)} Since, 3x > y, so, 3x - y, cannot be negative. So, 3x - y = 2 ....(iii) We know that, a³ - b³ = (a - b) × (a² + b² + ab). So, (3x)³ - (y)³ = 27x³ - y³ = (3x - y) × (9x² + y² + 3x × y) Putting the value of equations (i), (ii) and (iii) in the above equation, we get, = 2 × (100 + 3 × 16) = 2 × (100 + 48) = 2 × 148 = 296 Hence, option a.