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(D) The given equations are:$1$) $\frac{x}{a} + \frac{y}{b} - (a + b) = 0$$2$) $\frac{x}{a^2} + \frac{y}{b^2} - 2 = 0$Using the cross-multiplication m

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$(a^2, b^2)$

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(D) The given equations are:$1$) $\frac{x}{a} + \frac{y}{b} - (a + b) = 0$$2$) $\frac{x}{a^2} + \frac{y}{b^2} - 2 = 0$Using the cross-multiplication method for equations $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$:$\frac{x}{B_1C_2 - B_2C_1} = \frac{y}{C_1A_2 - C_2A_1} = \frac{1}{A_1B_2 - A_2B_1}$Here,$A_1 = \frac{1}{a}, B_1 = \frac{1}{b}, C_1 = -(a + b)$$A_2 = \frac{1}{a^2}, B_2 = \frac{1}{b^2}, C_2 = -2$$\frac{x}{(\frac{1}{b})(-2) - (\frac{1}{b^2})(-(a + b))} = \frac{y}{(-(a + b))(\frac{1}{a^2}) - (-2)(\frac{1}{a})} = \frac{1}{(\frac{1}{a})(\frac{1}{b^2}) - (\frac{1}{a^2})(\frac{1}{b})}$Simplifying the denominators:$\frac{x}{-\frac{2}{b} + \frac{a+b}{b^2}} = \frac{y}{-\frac{a+b}{a^2} + \frac{2}{a}} = \frac{1}{\frac{1}{ab^2} - \frac{1}{a^2b}}$$\frac{x}{\frac{-2b + a + b}{b^2}} = \frac{y}{\frac{-a - b + 2a}{a^2}} = \frac{1}{\frac{a - b}{a^2b^2}}$$\frac{x}{\frac{a - b}{b^2}} = \frac{y}{\frac{a - b}{a^2}} = \frac{a^2b^2}{a - b}$$x = \frac{a^2b^2}{a - b} \cdot \frac{a - b}{b^2} = a^2$$y = \frac{a^2b^2}{a - b} \cdot \frac{a - b}{a^2} = b^2$Thus,the solution is $(a^2, b^2)$.

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

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