If, 2x + y = 13, and 2xy = 12, and 2x > y, then find

Solution

2xy = 12 So, xy = (12/2) = 6 ......(i) 2x + y = 13 On squaring the above equation, we get, (2x + y)² = 13² Or, 4x² + y² + 2 × 2x × y = 169 Or, 4x² + y² + 4 × 6 = 169 (since, xy = 6) Or, 4x² + y² = 169 - 24 So, 4x² + y² = 145 ....(ii) (2x - y)² = 4x² + y² - 2 × 2x × y = 145 - 4 × 6 = 145 - 24 = 121 {using values from equations (i) and (ii)} Since, 2x > y, so, 2x - y, cannot be negative. So, 2x - y = 11 ....(iii) We know that, a³ - b³ = (a - b) × (a² + b² + ab). So, (2x)³ - (y)³ = 8x³ - y³ = (2x - y) × (4x² + y² + 2x × y) Putting the value of equations (i), (ii) and (iii) in the above equation, we get, = 11 × (145 + 2 × 6) = 11 × (145 + 12) = 11 × 157 = 1,727 Hence, option b.