# 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$

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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.

Updated on 10-Oct-2022 13:27:15