3 years ago

Please help will give brainist

Mathematics

1 answer:

Alexxx [7]3 years ago

5 0

Answer:

80 & 105

Step-by-step explanation:

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Paha777 [63]

$\bf \cfrac{n^2+3n+2}{n^2+6n+8}-\cfrac{2n}{n+4}\implies \cfrac{n^2+3n+21}{(n+4)(n+2)}-\cfrac{2n}{n+4}
\\\\\\
\textit{so our LCD will just be (n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(n+2)(2n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(2n^2+4n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~2n^2-4n}{(n+4)(n+2)}
\implies 
\cfrac{-n^2-n+2}{(n+4)(n+2)}
\\\\\\
\cfrac{-(n^2+n-2)}{(n+4)(n+2)}\implies \cfrac{-\underline{(n+2)}(n-1)}{(n+4)\underline{(n+2)}}\implies \cfrac{-(n-1)}{n+4}\implies \cfrac{1+n}{n+4}$