One of the solutions of the pair of equations a-x-b-y-0 and ab2-x-a2b-y-a2-b2 is-

We have nbsp; a/x b/y=0 ⇒a/x=b/y ⇒ x=a/by ...(i) and ab^2/x+ a^2b/y=a^2+b^2 ...(ii) Putting x=a/by in equation (ii), we get ab^2b/ay+a^2b/y=a^2+b^2 ⇒b/y( ...

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Standard VIII

Mathematics

Finding solution for consistent pair of LE

One of the so...

Question

One of the solutions of the pair of equations $\frac{a}{x} - \frac{b}{y} = 0 and \frac{a b^{2}}{x} + \frac{a^{2} b}{y} = a^{2} + b^{2} is:$

x = a

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y = - b

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x = - a

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x = a b^{2}

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Solution

The correct option is A $x = a$
We have
$\frac{a}{x} - \frac{b}{y} = 0 \Rightarrow \frac{a}{x} = \frac{b}{y} \Rightarrow x = \frac{a}{b} y . . . (i)$

$a n d \frac{a b^{2}}{x} + \frac{a^{2} b}{y} = a^{2} + b^{2} . . . (i i)$

$P u t t i n g x = \frac{a}{b} y i n e q u a t i o n (i i), w e g e t$

$\frac{a b^{2} b}{a y} + \frac{a^{2} b}{y} = a^{2} + b^{2} \Rightarrow \frac{b}{y} (b^{2} + a^{2}) = a^{2} + b^{2}$

$\Rightarrow y = b F r o m (i) x = \frac{a}{b} b = a$
$H e n c e x = a a n d y = b$

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One of the solutions of the pair of equations ax−by=0 and ab2x+a2by=a2+b2 is:

The correct option is A x=aWe have ax−by=0⇒ax=by⇒x=aby ...(i) and ab2x+a2by=a2+b2 ...(ii) Putting x=aby in equation (ii),we get ab2bay+a2by=a2+b2⇒by(b2+a2)=a2+b2 ⇒y=bFrom (i) x=abb=a Hence x=a and y=b

One of the solutions of the pair of equations $\frac{a}{x} - \frac{b}{y} = 0 and \frac{a b^{2}}{x} + \frac{a^{2} b}{y} = a^{2} + b^{2} is:$