The domination of fraction exceeds its numinator by 4. If the numinato

1.	The domination of fraction exceeds its numinator by 4. If the numinator and dinominator are both incered by 3, the new fraction become 4/5 find the original fraction. Plezz tell the answer
Answer» $\bf\blue{\underbrace{AnsweEr\implies \frac{13}{17}}}$ $\sf\large\underline{Let:}$ $\sf{\implies The\: numerator\:be\:x}$ $\sf{\implies The\: <klux>DENOMINATOR</klux>\:be\:y}$ $\sf{\implies The\: orginal\: fraction\:be\:\frac{x}{y}}$ $\sf\large\underline{To\: Find:}$ $\sf{\implies The\: orginal\: fraction=?}$ $\sf\large\underline{Solution:}$ By setting EQUATION and calculate the value of x and y than calculate orginal fraction after PUTTING the value of x and y] $\sf\small\underline{Given\:in\:case\:(i):}$ The denominator of fraction exceeds its numinator by 4] $\rm{\implies y=x+4-------(i)}$ $\sf\small\underline{Given\:in\:case\:(ii):}$ The numerator and denominator are both increased by 3, the new fraction become 4/5] $\tt{\implies \dfrac{x+3}{y+3}=\dfrac{4}{5}}$ $\rm{\implies 5(x+3)=4(y+3)}$ $\rm{\implies 5x+15=4y+12}$ $\rm{\implies 5x-4y=12-15}$ $\rm{\implies 5x-4y=-3-----(ii)}$ Putting the value of y=x+4 here] $\rm{\implies 5x-4(x+4)=-3}$ $\rm{\implies 5x-4x-16=-3}$ $\rm{\implies x=-3+16\implies x=13}$ Now putting the value of x=13 in EQ (i) $\rm{\implies y=x+4}$ $\rm{\implies y=13+4\implies y=17}$ $\sf\large{Hence,}$ $\sf{\implies The\: orginal\: fraction=\dfrac{x}{y}}$ $\sf{\implies The\: orginal\: fraction=\frac{13}{17}}$