Question

Solve the following pair of equations:

$\frac{9}{x}-\frac{4}{y}=8$, $\frac{13}{x}+\frac{7}{y}=101$

• A. $x=\frac{2}{3};y=\frac{4}{3}$
• B. $x=\frac{1}{4};y=\frac{1}{7}$
• C. $x=\frac{5}{4};y=\frac{2}{5}$
• D. $x=\frac{3}{2};y=\frac{6}{5}$

Answer 1

The correct option is B $x=\frac{1}{4};y=\frac{1}{7}$

The equation $\frac{9}{x}-\frac{4}{y}=8$ can be solved as:

$\frac{9}{x}-\frac{4}{y}=8$

$\Rightarrow \frac{9y-4x}{xy}=8$

$\Rightarrow 9y-4x=8xy.........\left(1\right)$

The equation $\frac{13}{x}+\frac{7}{y}=101$ can be solved as:

$\frac{13}{x}+\frac{7}{y}=101$

$\Rightarrow \frac{13y+7x}{xy}=101$

$\Rightarrow 13y+7x=101xy.........\left(2\right)$

Multiply the equation 1 by $7$ and equation 2 by $4$. Then we get the equations:

$63y-28x=56xy.........\left(3\right)$

$52y+28x=404xy.........\left(4\right)$

Add equations 3 and 4 to eliminate $x$, because the coefficients of $x$ are the opposite. So, we get

$(63y+52y)+(-28x+28x)=56xy+404xy$

i.e. $115y=460xy$

i.e. $115=460x$

i.e. $x=\frac{115}{460}=\frac{1}{4}$

Substituting this value of $x$ in the equation 1, we get

$9y-4\left(\frac{1}{4}\right)=8\left(\frac{1}{4}\right)y\Rightarrow 9y-1=2y\Rightarrow 9y-2y=1\Rightarrow 7y=1\Rightarrow y=\frac{1}{7}$

Hence, the solution of the equations is $x=\frac{1}{4},y=\frac{1}{7}$.