Question

Solve the following systems of equations using elimination method.

$\frac{5}{x}-\frac{4}{y}=-2,\frac{2}{x}+\frac{3}{y}=13$

• A. $(\frac{1}{2},\frac{1}{3})$
• B. $(\frac{1}{3},\frac{1}{3})$
• C. $(\frac{1}{2},\frac{1}{4})$
• D. None of these

Answer 1

The correct option is A $(\frac{1}{2},\frac{1}{3})$

Consider the given equations.

$\frac{5}{x}-\frac{4}{y}=-2$

$-4x+5y=-2xy$

$4x-5y=2xy$ $-------\left(1\right)$

$\frac{2}{x}+\frac{3}{y}=13$

$3x+2y=13xy$ $-------\left(2\right)$

On multiplying by $3$ in equation $\left(1\right)$ and multiplying by $4$ in equation $\left(2\right)$ and subtract, we get

$12x-15y=6xy$

$12x+8y=52xy$

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$-23y=-46xy$

$x=\frac{1}{2}$ $[$ put in $\left(2\right)]$

$3\times \frac{1}{2}+2y=13\times \frac{1}{2}\times y$

$\frac{3}{2}=\frac{13y-4y}{2}$

$\frac{3}{2}=\frac{9y}{2}$

$\frac{1}{2}=\frac{3y}{2}$

$y=\frac{1}{3}$

Hence, the value of $x$ is $\frac{1}{2}$ and the value of $y$ is $\frac{1}{3}$.