Question

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

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?

• A. $\frac{2}{7}$
• B. $\frac{3}{5}$
• C. $\frac{6}{7}$
• D. $\frac{1}{9}$

Answer 1

The correct option is B $\frac{3}{5}$

Let the fraction be $\frac{x}{y}$
As per the question, we have
$\frac{x+1}{y-1}=1$ ...(1)
$\frac{x}{y+1}=\frac{1}{2}$ ...(2)
From (1) will get
$\Rightarrow x+1=y-1$
$\Rightarrow x-y=-2$ .....(3)
From (2) will get
$\Rightarrow 2x=y+1$
$\Rightarrow 2x-y=1$ .....(4)
Using Elimination Method:
Subtracting (3) and (4) will get
$\Rightarrow x-y-2x+y=-2-1$
$\Rightarrow -x=-3$
$\Rightarrow x=3$
Now putting $x$ in (4)
$\Rightarrow 2\times 3-y=1$
$\Rightarrow 6-y=1$
$\Rightarrow -y=-5$
$\Rightarrow y=5$
Hence, $x=3,y=5$