Respuesta :
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
[tex]\frac{N+3}{D+2} = \frac{N}{D} +\frac{1}{3}[/tex]
Therefore,
[tex]\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}[/tex]
Which gives
[tex]\frac{N+3}{N+7} - \frac{N}{N+5} =\frac{1}{3}[/tex] and then we have
[tex]\frac{N+15}{N^2+12\cdot N+35} = \frac{1}{3}[/tex]
Therefore, we have
[tex]3\cdot N+45 ={N^2+12\cdot N+35}{[/tex] and
[tex]N^2 +9\cdot N-10 =0[/tex]
Which gives
[tex](N-1)\cdot (N+10) = 0[/tex]
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
[tex]\frac{1}{6} \hspace {0.2 cm}[/tex]
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:
[tex]\frac{x+3}{x+7} -\frac{x}{x+5} =\frac{1}{3}[/tex]
[tex]\frac{x^2+8x+15-(x^2+7x)}{x^2+12x+35} =\frac{1}{3}[/tex]
[tex]\frac{x^2+8x+15-x^2-7x}{x^2+12x+35} =\frac{1}{3}[/tex]
[tex]\frac{x+15}{x^2+12x+35} =\frac{1}{3}[/tex]
[tex]x^2+12x+35=3(x+15)\\\\x^2+12x+35=3x+45\\\\x^2+12x+35-3x-45=0\\\\x^2+9x-10[/tex]
we then factorize
[tex](x-1)(x+10)\\\\x-1 = 0\\\\x= 1[/tex]
we will then substitute x for 1 in the fraction [tex]\frac{x}{x+5}[/tex]
[tex]=\frac{1}{5+1}=\frac{1}{6}[/tex]
This is the original fraction.
