What is the domain of $(\frac{4}{a + 1}) - (\frac{5}{a}) = (\frac{20}{a^{2} + a})$ $(4/(a+1))-(5/a) = (20/(a^2+a))$ ?

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Answer 1 · 2015-10-26T03:26:33

$a = {- 25}$ $a={-25}$

$[1] \frac{4}{a + 1} - \frac{5}{a} = \frac{20}{a^{2} + a}$ $[1]" "4/(a+1)-5/a=20/(a^2+a)$

Combine $\frac{4}{a + 1}$ $4/(a+1)$ and $- \frac{5}{a}$ $-5/a$ by making them have the same denominator.

$[2] \frac{4 (a)}{a (a + 1)} - \frac{5 (a + 1)}{a (a + 1)} = \frac{20}{a^{2} + a}$ $[2]" "(4(a))/(a(a+1))-(5(a+1))/(a(a+1))=20/(a^2+a)$

$[3] \frac{4 (a) - 5 (a + 1)}{a (a + 1)} = \frac{20}{a^{2} + a}$ $[3]" "(4(a)-5(a+1))/(a(a+1))=20/(a^2+a)$

$[4] \frac{4 a - 5 a - 5}{a^{2} + a} = \frac{20}{a^{2} + a}$ $[4]" "(4a-5a-5)/(a^2+a)=20/(a^2+a)$

$[5] \frac{- a - 5}{a^{2} + a} = \frac{20}{a^{2} + a}$ $[5]" "(-a-5)/(a^2+a)=20/(a^2+a)$

Multiply both sides by $(a^{2} + a)$ $(a^2+a)$ to remove the denominator.

$[6]" "cancel(a^2+a)((-a-5)/cancel(a^2+a))=cancel(a^2+a)(20/cancel(a^2+a))$

$[7]" "-a-5=20$

$[8]" "-a=25$

$[9]" "color(blue)(a=-25)$

What is the domain of (4/(a+1))-(5/a) = (20/(a^2+a))(4a+1)−(5a)=(20a2+a)(4/(a+1))-(5/a) = (20/(a^2+a))?