#### 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 1/2 if we only add 1 to the denominator. What is the fraction?

#### Solution

Let the fraction be `x/y`

According to the given information,

`(x+1)/(y-1)=1` ⇒ x - y = -2 (1)

`x/(y+1) = 1/2` ⇒ 2x-y = 1 (2)

Subtracting equation (1) from equation (2), we obtain

x = 3 (3)

Substituting this value in equation (1), we obtain

3 - y = -2

-y = -5

y = 5

Hence the fraction is `3/5`

Is there an error in this question or solution?

Solution 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 1/2 if we only add 1 to the denominator. What is the fraction? Concept: Algebraic Methods of Solving a Pair of Linear Equations - Elimination Method.