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Q4.    Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :

                (ii) A fraction becomes \frac{1}{3} when 1 is subtracted from the numerator and it becomes \frac{1}{4} when 8 is added to its denominator. Find the fraction.

Answers (1)

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Let numerator of a fraction be x and the denominator is y.

Now, According to the question,

\frac{x-1}{y}=\frac{1}{3}

\Rightarrow 3(x-1)=y

\Rightarrow 3x-3=y

\Rightarrow 3x-y=3........(1)

Also,

\frac{x}{y+8}=\frac{1}{4}

\Rightarrow 4x=y+8

\Rightarrow 4x-y=8.........(2)

Now, Subtracting (1) from (2) we get,

4x-3x=8-3

\Rightarrow x=5

Putting this value in (2) we get,

4(5)-y=8

\Rightarrow y=20-8

\Rightarrow y=12

Hence, the fraction is 

\frac{x}{y}=\frac{5}{12}.

Posted by

Pankaj Sanodiya

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