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# Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1 / 2 if we

Q2.    Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\frac{1}{2 }$ if we only add 1 to the denominator. What is the fraction?

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Let the numerator of the fraction be x and denominator is y,

Now, According to the question,

$\frac{x+1}{y-1}=1$

$\Rightarrow x+1=y-1$

$\Rightarrow x-y=-2.........(1)$

Also,

$\frac{x}{y+1}=\frac{1}{2}$

$\Rightarrow 2x=y+1$

$\Rightarrow 2x-y=1..........(2)$

Now, Subtracting (1) from (2)  we get

$x=3$

Putting this value in (1)

$3-y=-2$

$\Rightarrow y=5$

Hence

$x=3\:and\:y=5$

And the fraction is

$\frac{3}{5}$.

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