What is (2k)/(3k+12)+(k+7)/(k^2+4k) ?

The solution to (2k)/(3k+12)+(k+7)/(k^2+4k) is (2k^2+3k+21)/(3k(k+4))

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simplify (2k)/(3k+12)+(k+7)/(k^2+4k)

simplify

\frac{2 k}{3 k + 12} + \frac{k + 7}{k ^{2} + 4 k}

\frac{2 k ^{2} + 3 k + 21}{3 k ( k + 4 )}

Solution steps

\frac{2 k}{3 k + 12} + \frac{k + 7}{k ^{2} + 4 k}

Least Common Multiplier of

3 k + 12, k^{2} + 4 k : 3 k (k + 4)

Adjust Fractions based on the LCM

= \frac{2 k ^{2}}{3 k ( k + 4 )} + \frac{( k + 7 ) \cdot 3}{3 k ( k + 4 )}

Apply the fraction rule:

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

= \frac{2 k ^{2} + ( k + 7 ) \cdot 3}{3 k ( k + 4 )}

Simplify

2 k^{2} + (k + 7) \cdot 3 : 2 k^{2} + 3 k + 21

= \frac{2 k ^{2} + 3 k + 21}{3 k ( k + 4 )}

3 r^{3} + 21 r^{2} - 54 r

y^{2} - 3 y + 4 = 0

6 x^{\frac{3}{2}} - 18 x^{\frac{1}{2}} + 12 x^{- \frac{1}{2}}

2 (\frac{20}{29}) (\frac{21}{29})

∣ x ∣ + 8 > 9

What is (2k)/(3k+12)+(k+7)/(k^2+4k) ?
The solution to (2k)/(3k+12)+(k+7)/(k^2+4k) is (2k^2+3k+21)/(3k(k+4))