A fraction becomes $\frac{1}{3}$ when $1$ is subtracted from the numerator, and it becomes $\frac{1}{4}$ when $8$ is added to its denominator. Formulate the pair of linear equations in two variables representing this problem. Solve the linear equations step-by-step using the Substitution Method. Find the required fraction and verify your answer.

Q: A fraction becomes $\frac{1}{3}$ when $1$ is subtracted from the numerator, and it becomes $\frac{1}{4}$ when $8$ is added to its denominator. Formulate the pair of linear equations in two variables representing this problem. Solve the linear equations step-by-step using the Substitution Method. Find the required fraction and verify your answer.

Step 1: Define variables and formulate the fraction\nLet the numerator of the fraction be $= x$\nLet the denominator of the fraction be $= y$\nTherefore, the required fraction is $=\frac{x}{y}$ Step 2: Set up equations based on given conditions Condition 1: When $1$ is subtracted from the numerator, the fraction becomes $\frac{1}{3}$. $$\frac{x - 1}{y} = \frac{1}{3}$$\nCross-multiplying: $$3(x - 1) = y$$ $$3x - 3 = y$$ $$3x - y = 3 \quad \text{--- (Equation 1)}$$ Condition 2: When $8$ is added to the denominator, the fraction becomes $\frac{1}{4}$. $$\frac{x}{y + 8} = \frac{1}{4}$$\nCross-multiplying: $$4x = y + 8$$ $$4x - y = 8 \quad \text{--- (Equation 2)}$$ --- Step 3: Solving by Substitution Method \nFrom Equation (1), express $y$ in terms of $x$: $$y = 3x - 3 \quad \text{--- (Equation 3)}$$ \nSubstitute the expression for $y$ from Equation (3) into Equation (2): $$4x - (3x - 3) = 8$$ $$4x - 3x + 3 = 8$$ $$x + 3 = 8$$ $$x = 8 - 3$$ $$x = 5$$ \nNow, substitute $x = 5$ into Equation (3) to find $y$: $$y = 3(5) - 3$$ $$y = 15 - 3 = 12$$ --- Step 4: Write the Fraction\nThe numerator $x = 5$ and the denominator $y = 12$.\nTherefore, the required fraction is $\frac{5}{12}$. --- Step 5: Verification Subtract $1$ from numerator: $\frac{5 - 1}{12} = \frac{4}{12} = \frac{1}{3}$ (Correct) Add $8$ to denominator: $\frac{5}{12 + 8} = \frac{5}{20} = \frac{1}{4}$ (Correct)

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Accepted Answer

Step 1: Define variables and formulate the fraction
Let the numerator of the fraction be $= x$
Let the denominator of the fraction be $= y$
Therefore, the required fraction is $=\frac{x}{y}$ Step 2: Set up equations based on given conditions Condition 1: When $1$ is subtracted from the numerator, the fraction becomes $\frac{1}{3}$. $$\frac{x - 1}{y} = \frac{1}{3}$$
Cross-multiplying: $$3(x - 1) = y$$ $$3x - 3 = y$$ $$3x - y = 3 \quad 	ext{--- (Equation 1)}$$ Condition 2: When $8$ is added to the denominator, the fraction becomes $\frac{1}{4}$. $$\frac{x}{y + 8} = \frac{1}{4}$$
Cross-multiplying: $$4x = y + 8$$ $$4x - y = 8 \quad 	ext{--- (Equation 2)}$$ --- Step 3: Solving by Substitution Method 
From Equation (1), express $y$ in terms of $x$: $$y = 3x - 3 \quad 	ext{--- (Equation 3)}$$ 
Substitute the expression for $y$ from Equation (3) into Equation (2): $$4x - (3x - 3) = 8$$ $$4x - 3x + 3 = 8$$ $$x + 3 = 8$$ $$x = 8 - 3$$ $$x = 5$$ 
Now, substitute $x = 5$ into Equation (3) to find $y$: $$y = 3(5) - 3$$ $$y = 15 - 3 = 12$$ --- Step 4: Write the Fraction
The numerator $x = 5$ and the denominator $y = 12$.
Therefore, the required fraction is $\frac{5}{12}$. --- Step 5: Verification Subtract $1$ from numerator: $\frac{5 - 1}{12} = \frac{4}{12} = \frac{1}{3}$ (Correct) Add $8$ to denominator: $\frac{5}{12 + 8} = \frac{5}{20} = \frac{1}{4}$ (Correct)