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(B) The given equations are:$ax + by + (-a + b) = 0$$bx - ay + (-a - b) = 0$Comparing with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y

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$(1, -1)$

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(B) The given equations are:$ax + by + (-a + b) = 0$$bx - ay + (-a - b) = 0$Comparing with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = a, b_1 = b, c_1 = -a + b$$a_2 = b, b_2 = -a, c_2 = -a - b$Using the cross-multiplication method:$\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}{b(-a - b) - (-a)(-a + b)} = \frac{y}{(-a + b)(b) - (-a - b)(a)} = \frac{1}{a(-a) - b(b)}$$\frac{x}{-ab - b^2 - (a^2 - ab)} = \frac{y}{-ab + b^2 - (-a^2 - ab)} = \frac{1}{-a^2 - b^2}$$\frac{x}{-ab - b^2 - a^2 + ab} = \frac{y}{-ab + b^2 + a^2 + ab} = \frac{1}{-a^2 - b^2}$$\frac{x}{-(a^2 + b^2)} = \frac{y}{a^2 + b^2} = \frac{1}{-(a^2 + b^2)}$Now,solving for $x$ and $y$:$x = \frac{-(a^2 + b^2)}{-(a^2 + b^2)} = 1$$y = \frac{a^2 + b^2}{-(a^2 + b^2)} = -1$Thus,the solution is $(x, y) = (1, -1)$.

Solve the following pair of equations by the method of cross multiplication:
$ax + by - a + b = 0$
$bx - ay - a - b = 0$

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Solve the following pair of equations by the method of cross multiplication:$ax + by - a + b = 0$$bx - ay - a - b = 0$