The denominator of a fraction is 6 more than | vedclass

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(A) Let the numerator of the fraction be $x$ and the denominator be $y$. According to the problem,the denominator is $6$ more than the numerator: $y =

Vedclass · Accepted Answer

$\frac{5}{11}$

Vedclass · Answer

(A) Let the numerator of the fraction be $x$ and the denominator be $y$. According to the problem,the denominator is $6$ more than the numerator: $y = x + 6$. If $1$ is added to the numerator and $3$ is subtracted from the denominator,the fraction becomes $\frac{3}{4}$: $\frac{x + 1}{y - 3} = \frac{3}{4}$. Substitute $y = x + 6$ into the second equation: $\frac{x + 1}{(x + 6) - 3} = \frac{3}{4}$. $\frac{x + 1}{x + 3} = \frac{3}{4}$. Cross-multiply: $4(x + 1) = 3(x + 3)$. $4x + 4 = 3x + 9$. $4x - 3x = 9 - 4$. $x = 5$. Now,find $y$: $y = x + 6 = 5 + 6 = 11$. Therefore,the fraction is $\frac{x}{y} = \frac{5}{11}$.

The denominator of a fraction is $6$ more than the numerator. If $1$ is added to the numerator and $3$ is subtracted from the denominator,the fraction becomes $\frac{3}{4}$. Find the fraction.

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