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(A) Comparing the given equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = 1, b_1 = 1, c_1 = -(a + b)

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$(a, b)$

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(A) Comparing the given equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = 1, b_1 = 1, c_1 = -(a + b) = -a - b$$a_2 = a, b_2 = -b, c_2 = -(a^2 - b^2) = -a^2 + b^2$Using the cross multiplication method:$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$Substituting the values:$\frac{x}{(1)(-a^2 + b^2) - (-b)(-a - b)} = \frac{y}{(-a - b)(a) - (-a^2 + b^2)(1)} = \frac{1}{(1)(-b) - (a)(1)}$Simplifying the denominators:$\frac{x}{-a^2 + b^2 - (ab + b^2)} = \frac{y}{-a^2 - ab + a^2 - b^2} = \frac{1}{-b - a}$$\frac{x}{-a^2 - ab} = \frac{y}{-b^2 - ab} = \frac{1}{-(a + b)}$Now,solving for $x$:$x = \frac{-a^2 - ab}{-(a + b)} = \frac{-a(a + b)}{-(a + b)} = a$Solving for $y$:$y = \frac{-b^2 - ab}{-(a + b)} = \frac{-b(b + a)}{-(a + b)} = b$Thus,the solution is $(x, y) = (a, b)$.

Solve the following pair of equations by the method of cross multiplication:
$x + y - (a + b) = 0$
$ax - by - (a^2 - b^2) = 0$

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Solve the following pair of equations by the method of cross multiplication:$x + y - (a + b) = 0$$ax - by - (a^2 - b^2) = 0$