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(D) First,rewrite the equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:$1$) $x - 3y + 5 = 0$$2$) $3x - 7y + 13 = 0$H

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$(-2, 1)$

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(D) First,rewrite the equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:$1$) $x - 3y + 5 = 0$$2$) $3x - 7y + 13 = 0$Here,$a_1 = 1, b_1 = -3, c_1 = 5$ and $a_2 = 3, b_2 = -7, c_2 = 13$.Using the cross-multiplication formula:$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$Substitute the values:$\frac{x}{(-3)(13) - (-7)(5)} = \frac{y}{(5)(3) - (13)(1)} = \frac{1}{(1)(-7) - (3)(-3)}$$\frac{x}{-39 + 35} = \frac{y}{15 - 13} = \frac{1}{-7 + 9}$$\frac{x}{-4} = \frac{y}{2} = \frac{1}{2}$Solving for $x$: $x = \frac{-4}{2} = -2$Solving for $y$: $y = \frac{2}{2} = 1$Thus,the solution is $(-2, 1)$.

Solve the following pairs of equations by the cross-multiplication method:
$x - 3y = -5$
$3x = 7y - 13$

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Solve the following pairs of equations by the cross-multiplication method:$x - 3y = -5$$3x = 7y - 13$