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

Answer:

original fraction = 1/6

Step-by-step 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:

                  $\frac{x+3}{x+7} -\frac{x}{x+5} =\frac{1}{3}$

                $\frac{x^2+8x+15-(x^2+7x)}{x^2+12x+35} =\frac{1}{3}$

                $\frac{x^2+8x+15-x^2-7x}{x^2+12x+35} =\frac{1}{3}$

                  $\frac{x+15}{x^2+12x+35} =\frac{1}{3}$

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

                 we then factorize

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

 we will then substitute x for 1 in the fraction $\frac{x}{x+5}$

          $=\frac{1}{5+1}=\frac{1}{6}$

This is the original fraction.

Answer 2

Answer:

The original fraction is 1/6

Step-by-step explanation:

Here we have a word problem as follows

Let the Denominator be D and

The Numerator be N

D = 5 + N

$\frac{N+3}{D+2} = \frac{N}{D} +\frac{1}{3}$

Therefore,

$\frac{N+3}{N+5+2} = \frac{N}{N+5} +\frac{1}{3}\Rightarrow \frac{N+3}{N+7} = \frac{N}{N+5} +\frac{1}{3}$

Which gives

$\frac{N+3}{N+7} - \frac{N}{N+5} =\frac{1}{3}$  and then we have

$\frac{N+15}{N^2+12\cdot N+35} = \frac{1}{3}$

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

$\frac{1}{6} \hspace {0.2 cm}$