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(C) Given equations are:$3x + y - 5 = 0$  --- $(1)$$5x + 3y - 3 = 0$  --- $(2)$Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we g

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$(3, -4)$

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(C) Given equations are:$3x + y - 5 = 0$  --- $(1)$$5x + 3y - 3 = 0$  --- $(2)$Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = 3, b_1 = 1, c_1 = -5$$a_2 = 5, b_2 = 3, c_2 = -3$Using the cross-multiplication formula:$\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)(-3) - (3)(-5)} = \frac{y}{(-5)(5) - (-3)(3)} = \frac{1}{(3)(3) - (5)(1)}$$\frac{x}{-3 + 15} = \frac{y}{-25 + 9} = \frac{1}{9 - 5}$$\frac{x}{12} = \frac{y}{-16} = \frac{1}{4}$For $x$: $\frac{x}{12} = \frac{1}{4} \implies x = 3$For $y$: $\frac{y}{-16} = \frac{1}{4} \implies y = -4$Thus,the solution is $(3, -4)$.

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

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