Question

solve the following pairs of equation by reducing them to a pair of linear equation​

Answer 1

$\sf\blue{Question:}$

$\sf{Solve:}$

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

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

_________________________________

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ values \ of \ x \ and \ y \ are }$

$\sf{4 \ and \ 5 \ respectively.}$

$\sf\orange{Given:}$

$\sf{The \ given \ equations \ are}$

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

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

$\sf\pink{To \ find:}$

$\sf{The \ values \ of \ x \ and \ y.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{The \ given \ equations \ are}$

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

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

$\sf{Substitute \ \frac{1}{x-1} \ by \ a \ and \ \frac{1}{y-2} \ by \ b}$

$\sf{in \ the \ given \ equations.}$

$\sf{\therefore{5a+b=2...(1)}}$

$\sf{\therefore{6a-3b=1...(2)}}$

$\sf{Multiply \ equation (1) \ by \ 3}$

$\sf{15a+3b=3...(3)}$

$\sf{Add \ equations \ (2) \ and \ (3)}$

$\sf{6a-3b=1}$

$\sf{+}$

$\sf{15a+3b=6}$

______________________

$\sf{21a=7}$

$\sf{\therefore{a=\frac{7}{21}}}$

$\boxed{\sf{a=\frac{1}{3}}}$

$\sf{Substitute \ a=\frac{1}{3} \ in \ equation (1)}$

$\sf{5\times\frac{1}{3}+b=2}$

$\sf{\frac{5}{3}+b=2}$

$\sf{b=2-\frac{5}{3}}$

$\sf{b=\frac{6-5}{3}}$

$\boxed{\sf{b=\frac{1}{3}}}$

$\sf{Resubstituting \ values \ of \ a \ and \ b.}$

$\sf{\therefore{\frac{1}{x-1}=\frac{1}{3}}}$

$\sf{\therefore{x-1=3}}$

$\sf{\therefore{x=3+1}}$

$\boxed{\sf{\therefore{x=4}}}$

$\sf{\therefore{\frac{1}{y-2}=\frac{1}{3}}}$

$\sf{\therefore{y-2=3}}$

$\sf{\therefore{y=3+2}}$

$\boxed{\sf{\therefore{y=5}}}$

$\sf\purple{\tt{\therefore{The \ values \ of \ x \ and \ y \ are }}}$

$\sf\purple{\tt{4 \ and \ 5 \ respectively.}}$