What is (a+3)/(a^2-1)+(a-1)/(2a+2)+(a-4)/(4a-4) ?

The solution to (a+3)/(a^2-1)+(a-1)/(2a+2)+(a-4)/(4a-4) is (3a^2-3a+10)/(4(a+1)(a-1))

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simplify (a+3)/(a^2-1)+(a-1)/(2a+2)+(a-4)/(4a-4)

simplify

\frac{a + 3}{a ^{2} - 1} + \frac{a - 1}{2 a + 2} + \frac{a - 4}{4 a - 4}

\frac{3 a ^{2} - 3 a + 10}{4 ( a + 1 ) ( a - 1 )}

Solution steps

\frac{a + 3}{a ^{2} - 1} + \frac{a - 1}{2 a + 2} + \frac{a - 4}{4 a - 4}

Least Common Multiplier of

a^{2} - 1, 2 a + 2, 4 a - 4 : 4 (a + 1) (a - 1)

Adjust Fractions based on the LCM

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

Apply the fraction rule:

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

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

Simplify

(a + 3) \cdot 4 + 2 (a - 1)^{2} + (a - 4) (a + 1) : 3 a^{2} - 3 a + 10

= \frac{3 a ^{2} - 3 a + 10}{4 ( a + 1 ) ( a - 1 )}

x^{2} + x - 6 \geq 0

- 2 x^{2} - 4 x + 4 = 0

(3 x + 8) (3 x + 8) = 0

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

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

What is (a+3)/(a^2-1)+(a-1)/(2a+2)+(a-4)/(4a-4) ?
The solution to (a+3)/(a^2-1)+(a-1)/(2a+2)+(a-4)/(4a-4) is (3a^2-3a+10)/(4(a+1)(a-1))