Answer:
[tex]\frac{4x^2-3x}{(x+9)(x-4)}[/tex]
Step-by-step explanation:
The sum of two rational expressions is done in the following way:
[tex]\frac{a}{b}+\frac{c}{d} = \frac{a*d + c*b}{b*d}[/tex]
In this case we have the following rational expressions
[tex]\frac{3x}{x+9} + \frac{x}{x-4}[/tex]
So:
[tex]a=3x\\d=(x+9)\\c=x\\d=(x-4)[/tex]
Therefore
[tex]\frac{3x}{x+9} + \frac{x}{x-4}=\frac{3x(x-4)+x(x+9)}{(x+9)(x-4)}[/tex]
simplifying we obtain:
[tex]\frac{3x(x-4)+x(x+9)}{(x+9)(x-4)}=\frac{3x^2-12x+x^2+9x}{(x+9)(x-4)}\\\\\frac{3x^2-12x+x^2+9x}{(x+9)(x-4)}=\frac{4x^2-3x}{(x+9)(x-4)}[/tex]
