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(C) Converting into the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$2x + y - 35 = 0$  ........ $(1)$$3x + 4y - 65 = 0$

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$(15, 5)$

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(C) Converting into the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$2x + y - 35 = 0$  ........ $(1)$$3x + 4y - 65 = 0$  ........ $(2)$Comparing with the standard form,we get:$a_1 = 2, b_1 = 1, c_1 = -35$$a_2 = 3, b_2 = 4, c_2 = -65$Using the cross-multiplication formula:$\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}{(1)(-65) - (4)(-35)} = \frac{y}{(-35)(3) - (-65)(2)} = \frac{1}{(2)(4) - (3)(1)}$$\frac{x}{-65 + 140} = \frac{y}{-105 + 130} = \frac{1}{8 - 3}$$\frac{x}{75} = \frac{y}{25} = \frac{1}{5}$Now,solving for $x$:$\frac{x}{75} = \frac{1}{5} \implies x = \frac{75}{5} = 15$Solving for $y$:$\frac{y}{25} = \frac{1}{5} \implies y = \frac{25}{5} = 5$Thus,the solution is $(x, y) = (15, 5)$.

Solve the following pair of equations by the method of cross-multiplication:
$2x + y = 35$
$3x + 4y = 65$

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Solve the following pair of equations by the method of cross-multiplication:$2x + y = 35$$3x + 4y = 65$