Discover new knowledge and insights with IDNLearn.com's extensive Q&A database. Get the information you need from our experts, who provide reliable and detailed answers to all your questions.
Sagot :
To simplify the expression [tex]\(\frac{2n}{n+4} + \frac{7}{n-1}\)[/tex], we need to find a common denominator. The common denominator for the fractions is [tex]\((n+4)(n-1)\)[/tex].
1. Rewrite each fraction with the common denominator:
[tex]\[ \frac{2n}{n+4} = \frac{2n(n-1)}{(n+4)(n-1)} \][/tex]
[tex]\[ \frac{7}{n-1} = \frac{7(n+4)}{(n+4)(n-1)} \][/tex]
2. Add the two fractions:
[tex]\[ \frac{2n(n-1)}{(n+4)(n-1)} + \frac{7(n+4)}{(n+4)(n-1)} = \frac{2n(n-1) + 7(n+4)}{(n+4)(n-1)} \][/tex]
3. Expand the numerators:
[tex]\[ 2n(n-1) = 2n^2 - 2n \][/tex]
[tex]\[ 7(n+4) = 7n + 28 \][/tex]
4. Combine the expanded numerators:
[tex]\[ 2n^2 - 2n + 7n + 28 = 2n^2 + 5n + 28 \][/tex]
5. Combine the fractions:
[tex]\[ \frac{2n^2 + 5n + 28}{(n+4)(n-1)} \][/tex]
Thus, the expression [tex]\(\frac{2n}{n+4} + \frac{7}{n-1}\)[/tex] simplifies to [tex]\(\frac{2n^2 + 5n + 28}{(n+4)(n-1)}\)[/tex].
The correct answer is:
A. [tex]\(\frac{2n^2 + 5n + 28}{(n+4)(n-1)}\)[/tex]
1. Rewrite each fraction with the common denominator:
[tex]\[ \frac{2n}{n+4} = \frac{2n(n-1)}{(n+4)(n-1)} \][/tex]
[tex]\[ \frac{7}{n-1} = \frac{7(n+4)}{(n+4)(n-1)} \][/tex]
2. Add the two fractions:
[tex]\[ \frac{2n(n-1)}{(n+4)(n-1)} + \frac{7(n+4)}{(n+4)(n-1)} = \frac{2n(n-1) + 7(n+4)}{(n+4)(n-1)} \][/tex]
3. Expand the numerators:
[tex]\[ 2n(n-1) = 2n^2 - 2n \][/tex]
[tex]\[ 7(n+4) = 7n + 28 \][/tex]
4. Combine the expanded numerators:
[tex]\[ 2n^2 - 2n + 7n + 28 = 2n^2 + 5n + 28 \][/tex]
5. Combine the fractions:
[tex]\[ \frac{2n^2 + 5n + 28}{(n+4)(n-1)} \][/tex]
Thus, the expression [tex]\(\frac{2n}{n+4} + \frac{7}{n-1}\)[/tex] simplifies to [tex]\(\frac{2n^2 + 5n + 28}{(n+4)(n-1)}\)[/tex].
The correct answer is:
A. [tex]\(\frac{2n^2 + 5n + 28}{(n+4)(n-1)}\)[/tex]
Your presence in our community is highly appreciated. Keep sharing your insights and solutions. Together, we can build a rich and valuable knowledge resource for everyone. Trust IDNLearn.com for all your queries. We appreciate your visit and hope to assist you again soon.