Q3(v):

Form the pair of linear equations for the following problems and find their solution by substitution method. (v) A fraction becomes $\frac{9}{11}$, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$. Find 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 2 is added to both the numerator and the denominator, the fraction becomes $\frac{9}{11}$.

Condition 2: If 3 is added to both the numerator and the denominator, the fraction becomes $\frac{5}{6}$.

To Find:

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

Step 1: Formulating the Linear Equations

Based on Condition 1: $\frac{x + 2}{y + 2} = \frac{9}{11}$

Cross-multiplying to simplify: $11(x + 2) = 9(y + 2)$

$11x + 22 = 9y + 18$

$11x - 9y = 18 - 22$

$11x - 9y = -4$ --- (Equation 1)

Based on Condition 2: $\frac{x + 3}{y + 3} = \frac{5}{6}$

Cross-multiplying to simplify: $6(x + 3) = 5(y + 3)$

$6x + 18 = 5y + 15$

$6x - 5y = 15 - 18$

$6x - 5y = -3$ --- (Equation 2)

Step 2: Solving by Substitution Method

From Equation 2, express $x$ in terms of $y$:

$6x = 5y - 3$

$x = \frac{5y - 3}{6}$ --- (Equation 3)

Step 3: Substituting Equation 3 into Equation 1

$11\left(\frac{5y - 3}{6}\right) - 9y = -4$

Multiply the entire equation by 6 to eliminate the denominator:

$11(5y - 3) - 54y = -24$

$55y - 33 - 54y = -24$

$y - 33 = -24$

$y = -24 + 33$

$y = 9$

Step 4: Finding the value of $x$

Substitute $y = 9$ into Equation 3:

$x = \frac{5(9) - 3}{6}$

$x = \frac{45 - 3}{6}$

$x = \frac{42}{6}$

$x = 7$

Step 5: Conclusion

The numerator $x$ is 7 and the denominator $y$ is 9. Therefore, the fraction is $\frac{7}{9}$.

Final Answer: The required fraction is $\frac{7}{9}$.