Question

If m 2

= 7x + 7, m 3=
4y, and m 4
= 112, find the values of x and y.

X = 112, y = 68
x = 15, y = 17
X = 17, y = 15
X = 68, y = 112

Answer 1

x = 1 and y = 4

Explanation:

m² = 7x + 7; m³= 4y and m∧4 = 112

√(m∧4) = √112

∴ m² = √112

Hence, 7x + 7 = √112

(7x + 7)² = 112

49x² + 14x + 49 = 112

49x² + 14x - 63 = 0

7x² + 2x - 9 = 0

7x² + 9x - 7x - 9 = 0

x(7x + 9) - 1(7x + 9) = 0

(x - 1)(7x + 9) = 0

x - 1 = 0

∴ x = 1

When x = 1

m²= 7 + 7 = 14

m³= 4y and m∧4 = 112

Also m∧4/m²= m² = 112/14 = 8

Hence, m° = 2; m = 2 X 2 = 4; m² = 2 x 2 x 2 = 8; m³= 2 x 2 x2 x 2 = 16

m³ = 16 = 4y

∴ y = 16/4 = 4