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(A) First,simplify the equations:$1$) Multiply the first equation by $20$: $4x - 5y = 4.5$ or $8x - 10y = 9$,so $8x - 10y - 9 = 0$.$2$) Multiply the s

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$(\frac{1}{2}, -\frac{1}{2})$

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(A) First,simplify the equations:$1$) Multiply the first equation by $20$: $4x - 5y = 4.5$ or $8x - 10y = 9$,so $8x - 10y - 9 = 0$.$2$) Multiply the second equation by $12$: $4x - 6y = 5$,so $4x - 6y - 5 = 0$.Using the cross-multiplication method for $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$Here,$a_1 = 8, b_1 = -10, c_1 = -9$ and $a_2 = 4, b_2 = -6, c_2 = -5$.$\frac{x}{(-10)(-5) - (-6)(-9)} = \frac{y}{(-9)(4) - (-5)(8)} = \frac{1}{(8)(-6) - (4)(-10)}$$\frac{x}{50 - 54} = \frac{y}{-36 + 40} = \frac{1}{-48 + 40}$$\frac{x}{-4} = \frac{y}{4} = \frac{1}{-8}$$x = \frac{-4}{-8} = \frac{1}{2}$ and $y = \frac{4}{-8} = -\frac{1}{2}$.Thus,the solution is $(\frac{1}{2}, -\frac{1}{2})$.

Solve the following pair of equations by the cross-multiplication method:
$\frac{x}{5} - \frac{y}{4} = \frac{9}{40}$
$\frac{x}{3} - \frac{y}{2} = \frac{5}{12}$

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Solve the following pair of equations by the cross-multiplication method:$\frac{x}{5} - \frac{y}{4} = \frac{9}{40}$$\frac{x}{3} - \frac{y}{2} = \frac{5}{12}$