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(B) Given equations are:$1$) $\frac{x}{a} - \frac{y}{b} = 0 \implies bx - ay = 0$$2$) $ax + by = a^2 + b^2 \implies ax + by - (a^2 + b^2) = 0$Using th

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$(a, b)$

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(B) Given equations are:$1$) $\frac{x}{a} - \frac{y}{b} = 0 \implies bx - ay = 0$$2$) $ax + by = a^2 + b^2 \implies ax + by - (a^2 + b^2) = 0$Using the cross-multiplication method for $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$Here,$a_1 = b, b_1 = -a, c_1 = 0$ and $a_2 = a, b_2 = b, c_2 = -(a^2 + b^2)$.$\frac{x}{(-a)(-(a^2 + b^2)) - (b)(0)} = \frac{y}{(0)(a) - (-(a^2 + b^2))(b)} = \frac{1}{(b)(b) - (a)(-a)}$$\frac{x}{a(a^2 + b^2)} = \frac{y}{b(a^2 + b^2)} = \frac{1}{b^2 + a^2}$$x = \frac{a(a^2 + b^2)}{a^2 + b^2} = a$$y = \frac{b(a^2 + b^2)}{a^2 + b^2} = b$Thus,the solution is $(a, b)$.

Solve the following pair of equations by the cross-multiplication method:
$\frac{x}{a} - \frac{y}{b} = 0$
$ax + by = a^2 + b^2$

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Solve the following pair of equations by the cross-multiplication method:$\frac{x}{a} - \frac{y}{b} = 0$$ax + by = a^2 + b^2$