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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?

Answers (1)

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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}.

Posted by

Pankaj Sanodiya

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