Question

The denominator of a fraction in its simplest form is greater than the numerator by 5 . If 3 is added to the numerator, and 2 added to the denominator, then the fraction itself is increased by 1/3 . Find the original fraction.

Answer 1

original fraction = 1/6

Explanation:

Since the denominator is greater than the numerator by 5, then the fraction should be:

x/x+5.

Again, if 3 is added to the numerator and 2 added to the denominator, then the fraction is increased by 1/3.

This means -

x+3/x+5+2

= x+3/x+7

As it is now, the fraction is already increased by one third of the original fraction.

That is:

`(x+3)/(x+7) -(x)/(x+5) =(1)/(3)`

`(x^2+8x+15-(x^2+7x))/(x^2+12x+35) =(1)/(3)`

`(x^2+8x+15-x^2-7x)/(x^2+12x+35) =(1)/(3)`

`(x+15)/(x^2+12x+35) =(1)/(3)`

`x^2+12x+35=3(x+15)\\\\x^2+12x+35=3x+45\\\\x^2+12x+35-3x-45=0\\\\x^2+9x-10`

we then factorize

`(x-1)(x+10)\\\\x-1 = 0\\\\x= 1`

we will then substitute x for 1 in the fraction
`(x)/(x+5)`

`=(1)/(5+1)=(1)/(6)`

This is the original fraction.

Answer 2

The original fraction is 1/6

Explanation:

Here we have a word problem as follows

Let the Denominator be D and

The Numerator be N

D = 5 + N

`(N+3)/(D+2) = (N)/(D) +(1)/(3)`

Therefore,

`(N+3)/(N+5+2) = (N)/(N+5) +(1)/(3)\Rightarrow (N+3)/(N+7) = (N)/(N+5) +(1)/(3)`

Which gives

`(N+3)/(N+7) - (N)/(N+5) =(1)/(3)` and then we have

`(N+15)/(N^2+12\cdot N+35) = (1)/(3)`

Therefore, we have

`3\cdot N+45 ={N^2+12\cdot N+35}{` and

`N^2 +9\cdot N-10 =0`

Which gives

`(N-1)\cdot (N+10) = 0`

That is the numerator = 1 or -10

and the denominator = N + 5 is therefore,

1 + 5 = 6 or

-10 + 5 = -5

The original fraction is therefore

`(1)/(6) \hspace {0.2 cm}`