Which of the following pairs of linear equations has a unique solution,no solution,or infinitely many solutions? In case there is a unique solution,find it by using the cross-multiplication method.
$2x + y = 5$
$3x + 2y = 8$

(A) Given equations are:
$2x + y - 5 = 0$
$3x + 2y - 8 = 0$
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = 2, b_1 = 1, c_1 = -5$
$a_2 = 3, b_2 = 2, c_2 = -8$
Calculating ratios:
$\frac{a_1}{a_2} = \frac{2}{3}, \quad \frac{b_1}{b_2} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-5}{-8} = \frac{5}{8}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$,the pair of equations has a unique solution.
Using the cross-multiplication method:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
$\frac{x}{(1)(-8) - (2)(-5)} = \frac{y}{(-5)(3) - (-8)(2)} = \frac{1}{(2)(2) - (3)(1)}$
$\frac{x}{-8 + 10} = \frac{y}{-15 + 16} = \frac{1}{4 - 3}$
$\frac{x}{2} = \frac{y}{1} = \frac{1}{1}$
Thus,$x = 2$ and $y = 1$.