What is (k+5)/(k+1)+8/(k^2-1) ?

The solution to (k+5)/(k+1)+8/(k^2-1) is (k+3)/(k-1)

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simplify (k+5)/(k+1)+8/(k^2-1)

simplify

\frac{k + 5}{k + 1} + \frac{8}{k ^{2} - 1}

\frac{k + 3}{k - 1}

Solution steps

\frac{k + 5}{k + 1} + \frac{8}{k ^{2} - 1}

Least Common Multiplier of

k + 1, k^{2} - 1 : (k + 1) (k - 1)

Adjust Fractions based on the LCM

= \frac{( k + 5 ) ( k - 1 )}{( k + 1 ) ( k - 1 )} + \frac{8}{( k + 1 ) ( k - 1 )}

Apply the fraction rule:

\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}

= \frac{( k + 5 ) ( k - 1 ) + 8}{( k + 1 ) ( k - 1 )}

Simplify

(k + 5) (k - 1) + 8 : k^{2} + 4 k + 3

= \frac{k ^{2} + 4 k + 3}{( k + 1 ) ( k - 1 )}

Cancel

\frac{k ^{2} + 4 k + 3}{( k + 1 ) ( k - 1 )} : \frac{k + 3}{k - 1}

= \frac{k + 3}{k - 1}

\frac{12 x ^{- 3} x ^{- 9} y ^{16}}{4 x ^{- 6} x ^{- 4} y ^{2}}

m - 2 < - 8

- 120 = 10 x

12 x^{2} - 15 x

\frac{2}{3} \cdot \frac{5}{4}

What is (k+5)/(k+1)+8/(k^2-1) ?
The solution to (k+5)/(k+1)+8/(k^2-1) is (k+3)/(k-1)