question_answer

The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original fraction is \[\frac{29}{20}\]. Find the original fraction.

Answer:

Let the denominator of the fraction be x then numerator is \[x-3\] and fraction is \[\frac{x-3}{x}\].

If 2 is added to both numerator and denominator then new fraction is

\[\frac{x-3+2}{x}=\frac{x-1}{x+2}\]

According to the question,

\[\frac{x-3}{x}+\frac{x-1}{x+2}=\frac{29}{20}\]

\[\Rightarrow \frac{(x-3)(x+2)+x(x-1)}{x(x+2)}=\frac{29}{20}\]

\[\Rightarrow \,20({{x}^{2}}-3x+2x-6+{{x}^{2}}-x)=29({{x}^{2}}+2x)\]

\[\Rightarrow 40{{x}^{2}}-40x-120=29{{x}^{2}}+58x\]

\[\Rightarrow 11{{x}^{2}}-98x-120=0\]

\[\Rightarrow 11{{x}^{2}}-110x+12x-120=0\]

\[\Rightarrow 11x(x-10)+12(x-10)=0\]

\[\Rightarrow (11x+12)(x-10)=0\]

\[x=10\] or \[-\frac{12}{11}\] (neglect)

Hence, the fraction is \[\frac{10-3}{10}\] i.e., \[\frac{7}{10}\].