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Solve by elimination method:$ax+by=1$ and $bx+ay=\frac{(a+b)^2}{a^2+b^2}-1$.
Given :
$ax+by=1$
$bx+ay=\frac{(a+b)^2}{a^2+b^2}-1$
To do:
We have to solve the given equations by elimination method.
Solution :
$ax+by=1$ .....................................(i)
$bx+ay=\frac{(a+b)^2}{a^2+b^2}-1$
Take L C M on Right hand side,
$\displaystyle bx\ +\ ay\ \ =\ \frac{( a+b)^{2} \ -\ \left( a^{2} \ +\ b^{2} \ \right)}{a^{2} \ +\ b^{2} \ }$
$\displaystyle bx\ +\ ay\ \ =\ \frac{a^{2} +b^{2} \ +\ 2ab\ -\ a^{2} \ -\ b^{2} \ }{a^{2} \ +\ b^{2} \ }$
$$\displaystyle ( a+b)^{2} \ =a^{2} +b^{2} \ +\ 2ab\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a^{2} \ -\ a^{2} \ =\ 0\ ;\ b^{2} \ -\ b^{2} \ =\ 0\ $$
$\displaystyle bx\ +\ ay\ \ =\ \frac{\ 2ab\ \ }{a^{2} \ +\ b^{2} \ }$......................(ii)
Multiply (i) by a,
$\displaystyle a^{2} x\ +\ aby\ =\ a$............................(iii)
Multiply (ii) by b,
$\displaystyle b^{2} x\ +\ aby\ \ =\ \frac{\ 2ab^{2} \ \ }{a^{2} \ +\ b^{2} \ }$..........(iv)
$\displaystyle ( iii) \ -\ ( iv) \ \ \Rightarrow $
$\displaystyle a^{2} x\ -\ b^{2} x\ +\ aby\ \ -\ aby\ =a\ -\ \ \frac{\ 2ab^{2} \ \ }{a^{2} \ +\ b^{2} \ }$
$\displaystyle \begin{array}{{>{\displaystyle}l}} a^{2} x\ -\ b^{2} x\ \ =a\ -\ \ \frac{\ 2ab^{2} \ \ }{a^{2} \ +\ b^{2} \ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ aby\ -\ aby\ =\ 0\ \ \end{array}$
$\displaystyle \begin{array}{{>{\displaystyle}l}} x\left( a^{2} \ -\ b^{2}\right) \ =a\ -\ \ \frac{\ 2ab^{2} \ \ }{a^{2} \ +\ b^{2} \ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$
Take L C M on Right Hand Side,
$ x\left( a^{2} \ -\ b^{2}\right) \ =\ \ \frac{a\ \left( a^{2} \ +\ b^{2} \ \right) \ -\ \ 2ab^{2} \ \ }{a^{2} \ +\ b^{2} \ }$
Take a as common on Right Hand Side
$x\left( a^{2} \ -\ b^{2}\right) \ =\ \ \frac{a\ \left[ a^{2} \ +\ b^{2} \ \ -\ 2b^{2}\right] \ \ }{a^{2} \ +\ b^{2} \ }$
$\displaystyle x\left( a^{2} \ -\ b^{2}\right) \ =\ \ \frac{a\ \left[ a^{2} -b^{2}\right] \ \ }{a^{2} \ +\ b^{2} \ } \ \ \ \ \ \ $
$x=\ \ \frac{a\ \left[ a^{2} -b^{2}\right] \ \ }{\left( a^{2} \ +\ b^{2}\right)\left( a^{2} \ -\ b^{2}\right) \ \ }$
$x\ =\ \frac{a}{a^{2} \ +\ b^{2}}$
Substitute 'x' value in (i),
$a x\ +\ by\ =\ 1$
$a\left( \ \frac{a}{a^{2} \ +\ b^{2}} \ \right) \ +\ by\ =\ 1$
$\frac{a^{2}}{a^{2} \ +\ b^{2}} \ +\ by\ =\ 1$
$by\ \ =\ \ 1\ -\ \frac{a^{2}}{a^{2} \ +\ b^{2}}$
Take LCM on Right Hand Side
$by\ \ =\ \ \ \frac{a^{2} \ +\ b^{2} -a^{2}}{a^{2} \ +\ b^{2}}$
$by\ \ =\ \ \ \frac{\ b^{2}}{a^{2} \ +\ b^{2}}$
$y\ \ =\ \ \ \frac{\ b^{2}}{b\left( a^{2} \ +\ b^{2}\right)}$
$y\ =\ \frac{\ b}{a^{2} \ +\ b^{2}}$
$x\ =\ \frac{a}{a^{2} \ +\ b^{2}}$.
$y\ =\ \frac{\ b}{a^{2} \ +\ b^{2}}$.
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