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(B) Comparing the given equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = 1, b_1 = 2, c_1 = 1$$a_2 =

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$3, -2$

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(B) Comparing the given equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = 1, b_1 = 2, c_1 = 1$$a_2 = 2, b_2 = -3, c_2 = -12$Using the cross-multiplication formula:$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$Substituting the values:$\frac{x}{(2)(-12) - (-3)(1)} = \frac{y}{(1)(2) - (-12)(1)} = \frac{1}{(1)(-3) - (2)(2)}$$\frac{x}{-24 + 3} = \frac{y}{2 + 12} = \frac{1}{-3 - 4}$$\frac{x}{-21} = \frac{y}{14} = \frac{1}{-7}$Now,solving for $x$:$\frac{x}{-21} = \frac{1}{-7} \implies x = \frac{-21}{-7} = 3$Solving for $y$:$\frac{y}{14} = \frac{1}{-7} \implies y = \frac{14}{-7} = -2$Thus,the solution is $x = 3, y = -2$.

Solve the following pair of equations by the method of cross-multiplication:
$x + 2y + 1 = 0$
$2x - 3y - 12 = 0$

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Solve the following pair of equations by the method of cross-multiplication:$x + 2y + 1 = 0$$2x - 3y - 12 = 0$