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., \[\frac{7}{10}\].
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