The sum is [tex]\frac{x+2}{x+3}[/tex].
Solution:
Given equation:
[tex]$\frac{x-2}{x+3}+\frac{10 x}{x^{2}-9}[/tex]
[tex]$=\frac{x-2}{x+3}+\frac{10 x}{x^{2}-3^2}[/tex]
Using algebraic expression: [tex]a^2-b^2=(a-b)(a+b)[/tex]
[tex]$=\frac{x-2}{x+3}+\frac{10 x}{(x-3)(x+3)}[/tex]
To make the denominator same, multiply and divide the first term by (x –3)
[tex]$=\frac{(x-2)(x-3)}{(x+3)(x-3)}+\frac{10 x}{(x-3)(x+3)}[/tex]
[tex]$=\frac{(x-2)(x-3)+10x}{(x+3)(x-3)}[/tex]
[tex]$=\frac{x^2-2x-3x+6+10x}{(x+3)(x-3)}[/tex]
[tex]$=\frac{x^2+5x+6}{(x+3)(x-3)}[/tex]
[tex]$=\frac{(x+3)(x+2)}{(x+3)(x-3)}[/tex]
Cancel the common factors.
[tex]$=\frac{x+2}{x+3}[/tex]
[tex]$\frac{x-2}{x+3}+\frac{10 x}{x^{2}-9}=\frac{x+2}{x+3}[/tex] ; but x ≠ –3
If you substitute x = –3, the value is indeterminate.
Hence the sum is [tex]\frac{x+2}{x+3}[/tex].
