Solve the following pair of equations by cross multiplication rule.

$\frac{x}{a} + \frac{y}{b} = 0, \frac{x}{a} - \frac{y}{b} = 4$

x = 2 a, y = - 2 b

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x = - 3 a, y = - b

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x = a, y = 2 b

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None of these

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Solution

The correct option is A $x = 2 a, y = - 2 b$
The equations are
$\Rightarrow \frac{x}{a} + \frac{y}{b} = 0$

$\Rightarrow \frac{x}{a} - \frac{y}{b} = 4$
$L e t \frac{1}{a} = u a n d \frac{1}{b} = v s o$

$\Rightarrow x u + y v = 0$ ......(1)
$\Rightarrow x u - y v - 4 = 0$ ......(2)

By Cross Multiplication Method,

$\Rightarrow \frac{u}{4 y - 0} = \frac{v}{- 4 x - 0} = \frac{1}{x y + x y}$

$\Rightarrow \frac{u}{4 y} = \frac{1}{2 x y} \Rightarrow u (2 x y) = 4 y$
$∵ u = \frac{1}{a}$
$\Rightarrow \frac{1}{a} = \frac{4 y}{2 x y} \Rightarrow \frac{1}{a} = \frac{2}{x} \Rightarrow x = 2 a$
$\Rightarrow \frac{v}{- 4 x} = \frac{1}{2 x y} \Rightarrow v (2 x y) = 4 x$
$∵ v = \frac{1}{b}$
$\Rightarrow \frac{1}{b} = \frac{- 4 x}{2 x y} \Rightarrow \frac{1}{b} = \frac{- 2}{y} \Rightarrow y = - 2 b$
$x = 2 a a n d y = - 2 b$

Suggest Corrections

Similar questions

(a + 2 b) x + (2 a - b) y = 2

(a - 2 b) x + (2 a + b) y = 3

(a + 2 b) x + (2 a - b) y = 2, (a - 2 b) x + (2 a + b) y = 3

(a + 2 b) x + (2 a - b) y = 2

(a - 2 b) x + (2 a + b) y = 3

(a + b) x + (a - b) y = 3 a - b (2 a^{2} + b^{2}) x - (a^{2} - b) y = b^{2} + 2 b

x + y = a + b, a x - b y = a^{2} - b^{2}

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