If 1 is added to the denominator of a fractio | vedclass

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(B) Let the fraction be $\frac{x}{y}$.According to the first condition,adding $1$ to the denominator gives $\frac{x}{y+1} = \frac{1}{2}$,which implies

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(B) Let the fraction be $\frac{x}{y}$.According to the first condition,adding $1$ to the denominator gives $\frac{x}{y+1} = \frac{1}{2}$,which implies $2x = y + 1$,or $2x - y = 1$ ........$(1)$.According to the second condition,adding $1$ to the numerator gives $\frac{x+1}{y} = 1$,which implies $x + 1 = y$,or $y - x = 1$ ........$(2)$.Adding equations $(1)$ and $(2)$:$(2x - y) + (y - x) = 1 + 1$$x = 2$.Substituting $x = 2$ into equation $(2)$:$y - 2 = 1$$y = 3$.Therefore,the required fraction is $\frac{x}{y} = \frac{2}{3}$.

If $1$ is added to the denominator of a fraction,the fraction becomes $\frac{1}{2}$. If $1$ is added to the numerator of the fraction,the fraction becomes $1$. The fraction is

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