If, 3x + 2y = 15, and 2xy = 12, and 3x > 2y, then find

Solution

2xy = 12 So, xy = (12/2) = 6 ......(i) 3x + 2y = 15 On squaring the above equation, we get, (3x + 2y)² = 15² Or, 9x² + 4y² + 2 × 3x × 2y = 225 Or, 9x² + 4y² + 12 × 6 = 225 (since, xy = 6) Or, 9x² + 4y² = 225 - 72 So, 9x² + 4y² = 153 ....(ii) (3x - 2y)² = 9x² + 4y² - 2 × 3x × 2y = 153 - 12 × 6 = 153 - 72 = 81 {using values from equations (i) and (ii)} Since, 3x > 2y, so, 3x - 2y, cannot be negative. So, 3x - 2y = 9 ....(iii) We know that, a³ - b³ = (a - b) × (a² + b² + ab). So, (3x)³ - (2y)³ = 27x³ - 8y³ = (3x - 2y) × (9x² + 4y² + 3x × 2y) Putting the value of equations (i), (ii) and (iii) in the above equation, we get, = 9 × (153 + 6 × 6) = 9 × (153 + 36) = 9 × 189 = 1,701 Hence, option a.