WAEC Further Mathematics 2014 Paper

$8^{x} \div \left( \frac{1}{4} \right)^{y} = 1$

$(2^{3})^{x} \div (2^{-2})^{y} = 2^{0}$

$2^{3x - (-2y)} = 2^{0}$

$\implies 3x + 2y = 0 \ldots (1)$

$\log_{2}(x - 2y) = 1$

$x - 2y = 2^{1} = 2 \ldots (2)$

Solving equations 1 and 2,

$x = \frac{1}{2}, y = \frac{-3}{4}$

$(x - y) = \frac{1}{2} - \frac{-3}{4} = \frac{5}{4}$