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(D) Converting the equations into the standard form $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$:From $\frac{x}{a} + \frac{y}{b} =

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$(a, b)$

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(D) Converting the equations into the standard form $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$:From $\frac{x}{a} + \frac{y}{b} = 2$,we get $bx + ay = 2ab$,so $bx + ay - 2ab = 0$ ... $(1)$From $ax - by = a^{2} - b^{2}$,we get $ax - by - (a^{2} - b^{2}) = 0$ ... $(2)$Comparing with the standard form:$a_{1} = b, b_{1} = a, c_{1} = -2ab$$a_{2} = a, b_{2} = -b, c_{2} = -(a^{2} - b^{2}) = b^{2} - a^{2}$Using the cross multiplication method:$\frac{x}{b_{1}c_{2} - b_{2}c_{1}} = \frac{y}{c_{1}a_{2} - c_{2}a_{1}} = \frac{1}{a_{1}b_{2} - a_{2}b_{1}}$$\frac{x}{a(b^{2} - a^{2}) - (-b)(-2ab)} = \frac{y}{(-2ab)(a) - (b^{2} - a^{2})(b)} = \frac{1}{b(-b) - a(a)}$$\frac{x}{ab^{2} - a^{3} - 2ab^{2}} = \frac{y}{-2a^{2}b - b^{3} + a^{2}b} = \frac{1}{-b^{2} - a^{2}}$$\frac{x}{-a^{3} - ab^{2}} = \frac{y}{-a^{2}b - b^{3}} = \frac{1}{-(a^{2} + b^{2})}$$\frac{x}{-a(a^{2} + b^{2})} = \frac{y}{-b(a^{2} + b^{2})} = \frac{1}{-(a^{2} + b^{2})}$Equating to the constant term:$x = \frac{-a(a^{2} + b^{2})}{-(a^{2} + b^{2})} = a$$y = \frac{-b(a^{2} + b^{2})}{-(a^{2} + b^{2})} = b$Thus,the solution is $(x, y) = (a, b)$.

Solve the following pair of equations by the method of cross multiplication:
$\frac{x}{a} + \frac{y}{b} = 2$
$ax - by = a^{2} - b^{2}$

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Solve the following pair of equations by the method of cross multiplication:$\frac{x}{a} + \frac{y}{b} = 2$$ax - by = a^{2} - b^{2}$