Question

Form the pair of linear equation in the following problem, and find its solution (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 `1/2` if we only add 1 to the denominator. What is the fraction?

Answer 1

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is `x/y`

If 1 is added to the numerator and 1 is subtracted from the denominator, the fraction becomes 1. Thus, we have

`(x + 1)/(x - y)=1`

⇒ x + 1 = y - 1

⇒ x - y = -2       ...(i)

If 1 is added to the denominator, the fraction becomes `1/2`. Thus, we have

`x/(y + 1)=1/2`

`x = 1/2 (y + 1)`

`x - y/2 = 1/2`         ...(ii)

By subtracting (2) from (1) we have

`x - y - x + y/2=-2 - 1/2`

⇒ `-y + y/2=-2 - 1/2`

⇒ `-1/2y = -5/2`

⇒ y = 5

Now, putting y = 5 in (ii), we have 

`x - 5/2 = 1/2`

`x = 1/2 + 5/2`

`x=6/2`

x = 3

Thus, x = 3 and y = 5

Hence, the required fraction = `3/5`