which expression is equivalent to \\(\\frac{2n}{n + 4} + \\frac{7}{n - 1}\\) if no denominator equals…

which expression is equivalent to \\(\\frac{2n}{n + 4} + \\frac{7}{n - 1}\\) if no denominator equals zero?\n\na. \\(\\frac{2n^2 + 5n + 28}{(n + 4)(n - 1)}\n\nb. \\(\\frac{2n^2 + 5n + 4}{(n + 4)(n - 1)}\n\nc. \\(\\frac{2n^2 + 6n + 28}{(n + 4)(n - 1)}\n\nd. \\(\\frac{2n^2 + 6n + 4}{(n + 4)(n - 1)}\\)

which expression is equivalent to \\(\\frac{2n}{n + 4} + \\frac{7}{n - 1}\\) if no denominator equals zero?\n\na. \\(\\frac{2n^2 + 5n + 28}{(n + 4)(n - 1)}\n\nb. \\(\\frac{2n^2 + 5n + 4}{(n + 4)(n - 1)}\n\nc. \\(\\frac{2n^2 + 6n + 28}{(n + 4)(n - 1)}\n\nd. \\(\\frac{2n^2 + 6n + 4}{(n + 4)(n - 1)}\\)

Answer

Explanation:

Step1: Find a common denominator

The denominators are (n + 4) and (n - 1), so the common denominator is ((n + 4)(n - 1)).

Step2: Rewrite each fraction with the common denominator

For (\frac{2n}{n + 4}), multiply numerator and denominator by (n - 1): (\frac{2n(n - 1)}{(n + 4)(n - 1)}) For (\frac{7}{n - 1}), multiply numerator and denominator by (n + 4): (\frac{7(n + 4)}{(n + 4)(n - 1)})

Step3: Add the fractions

[ \begin{align*} \frac{2n(n - 1)}{(n + 4)(n - 1)}+\frac{7(n + 4)}{(n + 4)(n - 1)}&=\frac{2n(n - 1)+7(n + 4)}{(n + 4)(n - 1)}\ &=\frac{2n^2-2n + 7n + 28}{(n + 4)(n - 1)}\ &=\frac{2n^2+5n + 28}{(n + 4)(n - 1)} \end{align*} ]

Answer:

A. (\frac{2n^2 + 5n + 28}{(n + 4)(n - 1)})