$\frac{2}{x^{2}-9}+\frac{1}{x^{2}+9 x+18}$ $\newline$ Which expression is equivalent to the sum for all x>-2 ? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{3}{2 x^{2}+9 x+9}$ $\newline$ (B) $\frac{2 x+13}{x^{2}+9 x+18}$ $\newline$ (C) $\frac{3}{x^{2}+3 x-18}$ $\newline$ (D) $\frac{3}{x^{2}+9 x+18}$

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Algebra 1

Compare linear and exponential growth

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\frac{2}{x^{2}-9}+\frac{1}{x^{2}+9 x+18}

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Which expression is equivalent to the sum for all

x>-2

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Choose

1

answer:

\newline

(A)

\frac{3}{2 x^{2}+9 x+9}

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(B)

\frac{2 x+13}{x^{2}+9 x+18}

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(C)

\frac{3}{x^{2}+3 x-18}

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(D)

\frac{3}{x^{2}+9 x+18}

Find Common Denominator: First, we need to simplify the sum of the two fractions. To do this, we will find a common denominator. $\newline$ The denominators are $x^2 - 9$ and $x^2 + 9x + 18$ . We can factor these quadratics to find potential common factors. $\newline$ $x^2 - 9$ can be factored into $(x - 3)(x + 3)$ . $\newline$ $x^2 + 9x + 18$ can be factored into $(x + 3)(x + 6)$ . $\newline$ The common denominator will be the product of $(x - 3)$ , $(x + 3)$ , and $(x + 6)$ .
Express Fractions: Now we will express each fraction with the common denominator. $\newline$ For the first fraction, $(2)/(x^2 - 9)$ , we need to multiply the numerator and denominator by $(x + 6)$ to have the common denominator. $\newline$ So, $(2)/(x^2 - 9)$ becomes $(2(x + 6))/((x - 3)(x + 3)(x + 6))$ . $\newline$ For the second fraction, $(1)/(x^2 + 9x + 18)$ , we need to multiply the numerator and denominator by $(x - 3)$ to have the common denominator. $\newline$ So, $(1)/(x^2 + 9x + 18)$ becomes $(1(x - 3))/((x - 3)(x + 3)(x + 6))$ .
Add Fractions: Next, we will add the two fractions with the common denominator. $\newline$ $(\frac{2(x + 6)}{(x - 3)(x + 3)(x + 6)} + \frac{1(x - 3)}{(x - 3)(x + 3)(x + 6)})$ becomes $(\frac{2x + 12 + x - 3}{(x - 3)(x + 3)(x + 6)})$ . $\newline$ Simplifying the numerator, we get $(\frac{3x + 9}{(x - 3)(x + 3)(x + 6)})$ .
Factor Numerator: We can further simplify the numerator by factoring out a $3$ . $\newline$ $(3x + 9)$ can be factored as $3(x + 3)$ . $\newline$ So, the expression becomes $\frac{3(x + 3)}{(x - 3)(x + 3)(x + 6)}$ .
Cancel Common Factor: We notice that $(x + 3)$ is a common factor in the numerator and the denominator, so we can cancel it out. $\newline$ The expression simplifies to $\frac{3}{(x - 3)(x + 6)}$ .
Rewrite Denominator: Finally, we rewrite the denominator as a single quadratic expression. $\newline$ The denominator $(x - 3)(x + 6)$ expands to $x^2 + 6x - 3x - 18$ , which simplifies to $x^2 + 3x - 18$ . $\newline$ So, the simplified expression is $\frac{3}{x^2 + 3x - 18}$ .