The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is `(29)/(20)dot` Find the original fraction.

Explanation

Let the denominator of the required fraction be x.
Then, its numerator `=(x-3)`.
So, the original fraction is `((x-3))/(x).`
`:." "((x-3)+2)/(x+2)+((3))/(x)=(29)/(20)`
`implies" "((x-1))/((x+2))+((x-3))/(x)=(29)/(20)`
`implies" "(x(x-1)+(x-3)(x+2))/((x+2)x)=(29)/(20)`
`implies" "((x^(2)-x)+(x^(2)-x-6))/(x^(2)+2x)=(29)/(20)`
`implies" "20(2x^(2)-2x-6)=29(x^(2)+2x)`
`implies" "40x^(2)-40x-120=29x^(2)+58x`
`implies" "11x^(2)-98x-120=0`
`implies" "11x^(2)-110x+12x-120=0`
`implies" "11x(x-10)+12(x-10)=0`
`implies" "(x-10)(11x+12)=0`
`implies" "x-10=0" or "11x+12=0`
`implies" "x=10" or "x=(-12)/(11)`
`implies" "x=10" "[because" denominator of a fraction is never negative"]`
`:." ""denominator = 10 and numerator"=(10-3)=7.`
Hence, the required fraction is `(7)/(10).`