Question

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

Answer 1

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

Step-by-step explanation:

The initial expression is

`(x-2)/(x+3)+(10x)/(x^2-9)`

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

`(x-2)/(x+3)+(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} ((x-2)(x-3))/((x+3)(x-3))+(10x)/((x+3)(x-3)) \\ \frac{(x-2)(x-3)+10x_{}}{(x+3)(x-3)} \end{gathered}`

Simplifying, we get

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

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