Question

Add these rational expressions and express your answer in simplest form. 2x2−9+3x+3(1 point)Responses2(x−3)(x+3)+3x+3the fraction with numerator 2 and denominator open paren x minus 3 close paren times open paren x plus 3 close paren plus the fraction with numerator 3 and denominator x plus 32+3(x+3)(x−3)(x+3)the fraction with numerator 2 plus 3 times open paren x plus 3 close paren and denominator open paren x minus 3 close paren times open paren x plus 3 close paren3x−7(x−3)(x+3)the fraction with numerator 3 x minus 7 and denominator open paren x minus 3 close paren times open paren x plus 3 close paren3x−1(x−3)(x+3)

Answer 1

To add the rational expressions given, let's first rewrite them clearly. We're tasked with adding the following:

\[
\frac{2}{(x - 3)(x + 3)} + \frac{3}{x + 3}
\]

To add these fractions, we need a common denominator. The denominators are \((x - 3)(x + 3)\) and \(x + 3\). The least common denominator (LCD) is:

\[
(x - 3)(x + 3)
\]

Now, let's rewrite the second fraction with the common denominator:

\[
\frac{3}{x + 3} = \frac{3(x - 3)}{(x + 3)(x - 3)} = \frac{3(x - 3)}{(x - 3)(x + 3)}
\]

Now we can add the two fractions:

\[
\frac{2}{(x - 3)(x + 3)} + \frac{3(x - 3)}{(x - 3)(x + 3)} = \frac{2 + 3(x - 3)}{(x - 3)(x + 3)}
\]

Now simplify the numerator:

\[
2 + 3(x - 3) = 2 + 3x - 9 = 3x - 7
\]

Thus, we have:

\[
\frac{3x - 7}{(x - 3)(x + 3)}
\]

This is already in simplest form since \(3x - 7\) does not factor further in a way that will cancel with the denominator. Therefore, the final result of the addition is:

\[
\frac{3x - 7}{(x - 3)(x + 3)}
\]