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.
$x - 3y - 7 = 0$
$3x - 3y - 15 = 0$

(A) Given equations are:
$x - 3y - 7 = 0$ ... $(1)$
$3x - 3y - 15 = 0$ ... $(2)$
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = 1, b_1 = -3, c_1 = -7$
$a_2 = 3, b_2 = -3, c_2 = -15$
Calculating ratios:
$\frac{a_1}{a_2} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-3}{-3} = 1, \quad \frac{c_1}{c_2} = \frac{-7}{-15} = \frac{7}{15}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (i.e.,$\frac{1}{3} \neq 1$),the system has a unique solution.
Using the cross-multiplication method:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
$\frac{x}{(-3)(-15) - (-3)(-7)} = \frac{y}{(-7)(3) - (-15)(1)} = \frac{1}{(1)(-3) - (3)(-3)}$
$\frac{x}{45 - 21} = \frac{y}{-21 + 15} = \frac{1}{-3 + 9}$
$\frac{x}{24} = \frac{y}{-6} = \frac{1}{6}$
For $x$: $\frac{x}{24} = \frac{1}{6} \implies x = \frac{24}{6} = 4$
For $y$: $\frac{y}{-6} = \frac{1}{6} \implies y = \frac{-6}{6} = -1$
Thus,the unique solution is $x = 4, y = -1$.