Solve the following pairs of equations:
$ \frac{x}{a}+\frac{y}{b}=a+b $
$ \frac{x}{a^{2}}+\frac{y}{b^{2}}=2, a, b ≠0 $
Given:
\( \frac{x}{a}+\frac{y}{b}=a+b \)
\( \frac{x}{a^{2}}+\frac{y}{b^{2}}=2, a, b ≠ 0 \)
To do:
We have to solve the given pairs of linear equations.
Solution:
$\frac{x}{a}+\frac{y}{b}=a+b$..........(i)
$\frac{x}{a^{2}}+\frac{y}{b^{2}}=2$.........(ii)
Multiplying (i) by $\frac{1}{a}$ and then subtracting from (ii), we get,
$[\frac{x}{a^{2}}+\frac{y}{b^{2}}]-[\frac{x}{a^{2}}+\frac{y}{a b}]=2-(1+\frac{b}{a})$
$y(\frac{1}{b^{2}}-\frac{1}{a b})=2-1-\frac{b}{a}$
$y(\frac{a-b}{a b^{2}})=1-\frac{b}{a}$
$=(\frac{a-b}{a})$
$y=\frac{a b^{2}}{a}$
$y=b^{2}$
This implies,
$\frac{x}{a^{2}}+\frac{b^{2}}{b^{2}}=2$
$\frac{x}{a^{2}}=2-1$
$\frac{x}{a^{2}}=1$
$x=a^{2}$
Hence, the required values of $x$ and $y$ are $a^{2}$ and $b^{2}$ respectively.
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