Question

Solve the following pair of equations by reducing them to a pair of linear equations
$\frac{5}{x-1}+\frac{1}{y-2}=2$ and $\frac{6}{x-1}-\frac{3}{y-2}=1$

Answer 1

First equation:
$\frac{5}{x-1}+\frac{1}{y-2}=2$
or, $\frac{5(y-2)+x-1}{(x-1)(y-2)}=2$
$\frac{5y-10+x-1}{xy-2x-y+2}=2$
= $x+5y-11=2xy-4x-2y+4$
or ,$x+5y-11+4x+2y-4=2xy$
or, $5x+7y-15=2xy$
second equation:
$\frac{6}{x-1}-\frac{3}{y-2}=1$
$\frac{6y-12-3x+3}{xy-2x-y+2}=1$
= $6y-3x-9=xy-2x-y+2$
$6y-3x-9+2x+y-2=xy$
or, $7y-x-11=xy$
Multiply the second equation by 2 and subtract first equation from resultant
$14y-2x-22=2xy$
$7y+5x-15=2xy$
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$7y-7x-7=0$
$y-x-1=0$
$y=x+1$
Substituting the value of x in first equation, we get;
$5x+7y-15=2xy$
Or, $5x+7x+7-15-2x(x+1)$ Or, $12x-8=2x^2+2x$ Or, $10x-8=2x^2$ Or, $x^2=5x-4$
Similarly, substituting the value of $y$ in second equation, we get;
$7y-x-11=xy$
Or, $7x+7-x-11=x^2+1$ Or, $6x-5=x^2$
From above two equations, it is clear
$5x-4=6x-5$
Or, $5x+1=6x$
Or, $x=1$
Hence, $y=x+1=2$
Hence, $x=1$ and $y=2$