What is the original fraction if the denominator is 2 less than twice the numerator?

A. [tex]\(\frac{n}{2n-2}\)[/tex]

B. [tex]\(\frac{n}{2}\)[/tex]

C. [tex]\(\frac{n-2}{2}\)[/tex]

D. [tex]\(n\)[/tex]



Answer :

To find the original fraction where the denominator is 2 less than twice the numerator, let's denote the numerator by [tex]\( n \)[/tex].

1. According to the problem, the denominator is described as "2 less than twice the numerator."
- This can be written mathematically as: [tex]\( 2n - 2 \)[/tex].

2. Hence, the original fraction is:
[tex]\[ \frac{n}{2n - 2} \][/tex]

3. Among the given choices:
[tex]\[ \frac{n}{2n-2}, \quad \frac{n}{2}, \quad \frac{n-2}{2}, \quad n \][/tex]
the correct answer is clearly:
[tex]\[ \frac{n}{2n - 2} \][/tex]

Therefore, the original fraction is [tex]\(\frac{n}{2n - 2}\)[/tex].