What is (5n+5)/(5n^2+35n-40)+(7n)/(3n) ?

The solution to (5n+5)/(5n^2+35n-40)+(7n)/(3n) is (7n^2+52n-53)/(3(n-1)(n+8))

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simplify (5n+5)/(5n^2+35n-40)+(7n)/(3n)

simplify

\frac{5 n + 5}{5 n ^{2} + 35 n - 40} + \frac{7 n}{3 n}

\frac{7 n ^{2} + 52 n - 53}{3 ( n - 1 ) ( n + 8 )}

Solution steps

\frac{5 n + 5}{5 n ^{2} + 35 n - 40} + \frac{7 n}{3 n}

Cancel

\frac{5 n + 5}{5 n ^{2} + 35 n - 40} : \frac{n + 1}{( n - 1 ) ( n + 8 )}

= \frac{n + 1}{( n - 1 ) ( n + 8 )} + \frac{7 n}{3 n}

Cancel the common factor:

n

= \frac{n + 1}{( n - 1 ) ( n + 8 )} + \frac{7}{3}

Simplify

\frac{n + 1}{( n - 1 ) ( n + 8 )} + \frac{7}{3} : \frac{( n + 1 ) \cdot 3 + 7 ( n - 1 ) ( n + 8 )}{3 ( n - 1 ) ( n + 8 )}

= \frac{( n + 1 ) \cdot 3 + 7 ( n - 1 ) ( n + 8 )}{3 ( n - 1 ) ( n + 8 )}

Simplify

(n + 1) \cdot 3 + 7 (n - 1) (n + 8) : 7 n^{2} + 52 n - 53

= \frac{7 n ^{2} + 52 n - 53}{3 ( n - 1 ) ( n + 8 )}

4 - 2 n = 34

\frac{1}{4} - \frac{1}{2}

\frac{1}{4} \cdot 32

- 13 = - 6 (- 5) - 9

10 = 0.24 (y - 350)^{2}

What is (5n+5)/(5n^2+35n-40)+(7n)/(3n) ?
The solution to (5n+5)/(5n^2+35n-40)+(7n)/(3n) is (7n^2+52n-53)/(3(n-1)(n+8))