Question

which expression is equivalent to \\(\frac{2n}{n + 4} + \frac{7}{n - 1}\\) if no denominator equals zero?\
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a. \\(\frac{2n^2 + 5n + 28}{(n + 4)(n - 1)}\\)\
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b. \\(\frac{2n^2 + 5n + 4}{(n + 4)(n - 1)}\\)\
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c. \\(\frac{2n^2 + 6n + 28}{(n + 4)(n - 1)}\\)\
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d. \\(\frac{2n^2 + 6n + 4}{(n + 4)(n - 1)}\\)

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 \(n - 1\): \(\frac{2n(n - 1)}{(n + 4)(n - 1)}=\frac{2n^2-2n}{(n + 4)(n - 1)}\)
For \(\frac{7}{n - 1}\), multiply numerator and denominator by \(n + 4\): \(\frac{7(n + 4)}{(n + 4)(n - 1)}=\frac{7n + 28}{(n + 4)(n - 1)}\)

Step3: Add the two fractions

\(\frac{2n^2-2n}{(n + 4)(n - 1)}+\frac{7n + 28}{(n + 4)(n - 1)}=\frac{2n^2-2n + 7n + 28}{(n + 4)(n - 1)}\)
Simplify the numerator: \(2n^2+5n + 28\)
So the equivalent expression is \(\frac{2n^2+5n + 28}{(n + 4)(n - 1)}\)

Answer:

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