How do you simplify $\frac{a^{2}}{a - 3} = \frac{9}{a - 3}$ $(a^2)/(a-3)= 9/ (a-3)$ ?

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How do you simplify $\frac{a^{2}}{a - 3} = \frac{9}{a - 3}$ $(a^2)/(a-3)= 9/ (a-3)$ ?

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Answer 1 · 2016-04-12T23:10:49

Since everything is on the same denominator, we can eliminate the denominators and solve.

Explanation:

$a^{2} = 9$ $a^2 = 9$

$a = \sqrt{9}$ $a = sqrt(9)$

$a = \pm 3$ $a = +-3$

However, +3 would make the denominator equal to zero, which is undefined in the real number system. Therefore, the solution set is ${x = - 3}$ ${x = -3}$

Hopefully this helps!

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How do you simplify $\frac{a^{2}}{a - 3} = \frac{9}{a - 3}$ $(a^2)/(a-3)= 9/ (a-3)$ ?

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How do you simplify $\frac{a^{2}}{a - 3} = \frac{9}{a - 3}$ $(a^2)/(a-3)= 9/ (a-3)$ ?