Q2(i):

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?

Given:

A fraction where the numerator and denominator are unknown. Let the numerator be $x$ and the denominator be $y$. The fraction is represented as $\frac{x}{y}$.

Condition 1: If we add 1 to the numerator and subtract 1 from the denominator, the fraction reduces to 1.

Condition 2: If we only add 1 to the denominator, the fraction becomes $\frac{1}{2}$.

To Find:

The value of the fraction $\frac{x}{y}$ using the elimination method.

Step 1: Formulating the Linear Equations

Based on Condition 1: $\frac{x + 1}{y - 1} = 1$

Multiplying both sides by $(y - 1)$: $x + 1 = y - 1$

Rearranging the terms to standard form $ax + by = c$: $x - y = -2$ --- (Equation 1)

Based on Condition 2: $\frac{x}{y + 1} = \frac{1}{2}$

Using cross-multiplication: $2x = 1(y + 1)$

Rearranging the terms: $2x - y = 1$ --- (Equation 2)

Step 2: Applying the Elimination Method

We have the system of equations:

(1) $x - y = -2$

(2) $2x - y = 1$

To eliminate the variable $y$, we subtract Equation 1 from Equation 2:

$(2x - y) - (x - y) = 1 - (-2)$

$2x - y - x + y = 1 + 2$ [Distributing the negative sign]

$x = 3$ [Combining like terms]

Step 3: Solving for the second variable

Substitute $x = 3$ into Equation 1:

$3 - y = -2$

$-y = -2 - 3$ [Subtracting 3 from both sides]

$-y = -5$

$y = 5$ [Multiplying both sides by -1]

Step 4: Verification

Check with Condition 1: $\frac{3+1}{5-1} = \frac{4}{4} = 1$. (Correct)

Check with Condition 2: $\frac{3}{5+1} = \frac{3}{6} = \frac{1}{2}$. (Correct)

Final Answer: The fraction is $\frac{3}{5}$.