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(A) Given pair of linear equations is:$\frac{x}{a} + \frac{y}{b} = a + b \quad .....(i)$$\frac{x}{a^{2}} + \frac{y}{b^{2}} = 2, \quad a, b 
eq 0 \qua

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$a^{2}, b^{2}$

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(A) Given pair of linear equations is:$\frac{x}{a} + \frac{y}{b} = a + b \quad .....(i)$$\frac{x}{a^{2}} + \frac{y}{b^{2}} = 2, \quad a, b 
eq 0 \quad .....(ii)$Multiply equation $(i)$ by $\frac{1}{a}$:$\frac{x}{a^{2}} + \frac{y}{ab} = 1 + \frac{b}{a} \quad .....(iii)$Subtract equation $(iii)$ from equation $(ii)$:$\left( \frac{x}{a^{2}} + \frac{y}{b^{2}} ight) - \left( \frac{x}{a^{2}} + \frac{y}{ab} ight) = 2 - \left( 1 + \frac{b}{a} ight)$$\frac{y}{b^{2}} - \frac{y}{ab} = 1 - \frac{b}{a}$$y \left( \frac{a - b}{ab^{2}} ight) = \frac{a - b}{a}$Since $a 
eq b$ (assuming general case),we can cancel $(a - b)$:$\frac{y}{ab^{2}} = \frac{1}{a}$$y = b^{2}$Substitute $y = b^{2}$ into equation $(ii)$:$\frac{x}{a^{2}} + \frac{b^{2}}{b^{2}} = 2$$\frac{x}{a^{2}} + 1 = 2$$\frac{x}{a^{2}} = 1$$x = a^{2}$Thus,the solution is $x = a^{2}$ and $y = b^{2}$.

Solve the following pair of linear equations:
$\frac{x}{a} + \frac{y}{b} = a + b$
$\frac{x}{a^{2}} + \frac{y}{b^{2}} = 2, \quad a, b \neq 0$

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Solve the following pair of linear equations:$\frac{x}{a} + \frac{y}{b} = a + b$$\frac{x}{a^{2}} + \frac{y}{b^{2}} = 2, \quad a, b \neq 0$