How do you evaluate $\frac{a^{2} - 2 (b^{2} - a)}{2 + a}$ $(a^2 - 2(b^2-a)) /(2 +a)$ for a = 3 and b= 5?

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How do you evaluate $\frac{a^{2} - 2 (b^{2} - a)}{2 + a}$ $(a^2 - 2(b^2-a)) /(2 +a)$ for a = 3 and b= 5?

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Answer 1 · 2016-08-07T11:28:24

$= - 7$ $=-7$

Explanation:

$\frac{a^{2} - 2 (b^{2} - a)}{2 + a}$ $(a^2-2(b^2-a))/(2+a)$ for $a = 3$ $a=3$ and $b = 5$ $b=5$
$= \frac{3^{2} - 2 (5^{2} - 3)}{2 + 3}$ $=(3^2-2(5^2-3))/(2+3)$
$= \frac{9 - 2 (25 - 3)}{5}$ $=(9-2(25-3))/5$
$= \frac{9 - 2 (22)}{5}$ $=(9-2(22))/5$
$= \frac{9 - 44}{5}$ $=(9-44)/5$
$= - \frac{35}{5}$ $=-35/5$
$= - 7$ $=-7$

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How do you evaluate $\frac{a^{2} - 2 (b^{2} - a)}{2 + a}$ $(a^2 - 2(b^2-a)) /(2 +a)$ for a = 3 and b= 5?

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How do you evaluate $\frac{a^{2} - 2 (b^{2} - a)}{2 + a}$ $(a^2 - 2(b^2-a)) /(2 +a)$ for a = 3 and b= 5?