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(A) The given equations are:$ax - by - \frac{a-b}{2} = 0$  --- $(1)$$x + 3y - 2 = 0$  --- $(2)$Using the cross-multiplication method for $a_1x + b_1y

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$(\frac{1}{2}, \frac{1}{2})$

Vedclass · Answer

(A) The given equations are:$ax - by - \frac{a-b}{2} = 0$  --- $(1)$$x + 3y - 2 = 0$  --- $(2)$Using the cross-multiplication method for $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$Here,$a_1 = a, b_1 = -b, c_1 = -\frac{a-b}{2}$ and $a_2 = 1, b_2 = 3, c_2 = -2$.$\frac{x}{(-b)(-2) - (3)(-\frac{a-b}{2})} = \frac{y}{(-\frac{a-b}{2})(1) - (-2)(a)} = \frac{1}{(a)(3) - (1)(-b)}$$\frac{x}{2b + \frac{3a-3b}{2}} = \frac{y}{-\frac{a}{2} + \frac{b}{2} + 2a} = \frac{1}{3a + b}$$\frac{x}{\frac{4b + 3a - 3b}{2}} = \frac{y}{\frac{-a + b + 4a}{2}} = \frac{1}{3a + b}$$\frac{x}{\frac{3a + b}{2}} = \frac{y}{\frac{3a + b}{2}} = \frac{1}{3a + b}$Equating with the constant term:$x = \frac{3a+b}{2} \cdot \frac{1}{3a+b} = \frac{1}{2}$$y = \frac{3a+b}{2} \cdot \frac{1}{3a+b} = \frac{1}{2}$Thus,the solution is $(\frac{1}{2}, \frac{1}{2})$.

Solve the following pair of equations by the cross-multiplication method:
$ax - by = \frac{a-b}{2}$
$x + 3y = 2$

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Solve the following pair of equations by the cross-multiplication method:$ax - by = \frac{a-b}{2}$$x + 3y = 2$