Question

show your worksimplify the sum. State any restrictions on the variables.(x-2)/(x+3) + 10x/(x^2-9)

Answer 1

Answer:

(x + 2)/(x - 3) when x is different to 3 and -3.

Explanation:

The initial expression is

$\frac{x-2}{x+3}+\frac{10x}{x^2-9}$

Since (x² - 9) = (x + 3)(x - 3), we get:

$\frac{x-2}{x+3}+\frac{10x}{(x+3)(x-3)}$

Then, the denominator can't be zero, so x should be different to 3 and to -3. Because

x + 3 = 0 --> x = -3

or

x - 3 = 0 --> x = 3

Then, we can add the fractions as follows

$\begin{gathered} \frac{(x-2)(x-3)}{(x+3)(x-3)}+\frac{10x}{(x+3)(x-3)} \\ \frac{(x-2)(x-3)+10x_{}}{(x+3)(x-3)} \end{gathered}$

Simplifying, we get

$\begin{gathered} \frac{(x^2-5x+6)+10x}{(x+3)(x-3)} \\ =\frac{x^2+5x+6}{(x+3)(x-3)} \\ =\frac{(x+2)(x+3)_{}}{(x+3)(x-3_{})}=\frac{(x+2)}{(x-3)} \end{gathered}$

Therefore, the sum is equal to (x + 2)/(x - 3) when x is different to 3 and -3.