Solve the following pair of linear equations by the substitution and cross-multiplication methods:
$8x + 5y = 9$
$3x + 2y = 4$

(A) Given equations are:
$8x + 5y = 9$ $...(i)$
$3x + 2y = 4$ $...(ii)$
Substitution Method:
From equation $(ii)$,we get:
$3x = 4 - 2y \implies x = \frac{4 - 2y}{3}$ $...(iii)$
Substituting $(iii)$ in $(i)$:
$8(\frac{4 - 2y}{3}) + 5y = 9$
$32 - 16y + 15y = 27$
$-y = 27 - 32$
$-y = -5 \implies y = 5$
Substituting $y = 5$ in $(iii)$:
$x = \frac{4 - 2(5)}{3} = \frac{4 - 10}{3} = \frac{-6}{3} = -2$
Thus,$x = -2, y = 5$.
Cross-multiplication Method:
$8x + 5y - 9 = 0$
$3x + 2y - 4 = 0$
Using the formula $\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$:
$\frac{x}{(5)(-4) - (2)(-9)} = \frac{y}{(-9)(3) - (-4)(8)} = \frac{1}{(8)(2) - (3)(5)}$
$\frac{x}{-20 + 18} = \frac{y}{-27 + 32} = \frac{1}{16 - 15}$
$\frac{x}{-2} = \frac{y}{5} = \frac{1}{1}$
$x = -2, y = 5$.