Question

Solve the following pairs of equations by reducing them to a pair of linear equations:
$\frac{5}{x-1}+\frac{1}{y-2}=2$

$\frac{6}{x-1}-\frac{3}{y-2}=1$

Answer 1

The set of equation is given as,

$\frac{5}{x-1}+\frac{1}{y-2}=2$

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

$5y-10+x-1=2\left(xy-2x-y+2\right)$

$5y+x-11=2xy-4x-2y+4$

$5y+x-11+4x+2y-4-2xy=0$

$5x+7y-2xy=15$ (1)

And,

$\frac{6}{x-1}-\frac{3}{y-2}=1$

$6\left(y-2\right)-3\left(x-1\right)=1\left(x-1\right)\left(y-2\right)$

$6y-12-3x+3=xy-2x-y+2$

$6y-3x-9-xy+2x+y-2=0$

$-x+7y-xy-11=0$

$x-7y+xy+11=0$

$x-7y+xy=-11$ (2)

Adding equation (1) and equation(2),

$6x-xy=4$

$x\left(6-y\right)=4$

$x=\frac{4}{6-y}$

Substituting the value of $x$ in equation (1),

$5\left(\frac{4}{6-y}\right)+7y-2y\left(\frac{4}{6-y}\right)=15$

$\frac{20}{6-y}+7y-\frac{8y}{6-y}=15$

$20+7y\left(6-y\right)-8y=15\left(6-y\right)$

$20+42y-7y^2-8y=90-15y$

$7y^2-49y+70=0$

$y^2-7y+10=0$

$y^2-5y-2y+10=0$

$y\left(y-5\right)-2\left(y-5\right)=0$

$\left(y-2\right)\left(y-5\right)=0$

$y=2,5$

At $y=2$, the equation is undefined, so the only value of $y$ is 5.

Now, substitute the value of $y$ in $x=\frac{4}{6-y}$.

$x=\frac{4}{6-5}$

$x=\frac{4}{1}$

$x=4$

Therefore, the values are $x=4$ and $y=5$.