Answer:
[tex]\frac{x^2+8x+9}{(x+3)(x-1)}[/tex]
Step-by-step explanation:
[tex]\frac{2x}{x^2+2x-3}[/tex] + [tex]\frac{x+3}{x-1}[/tex] ← factor the denominator of first fraction
= [tex]\frac{2x}{(x+3)(x-1)}[/tex] + [tex]\frac{x+3}{x-1}[/tex]
Before adding we require the denominators to be the same
Multiply numerator/denominator of second fraction by (x + 3)
= [tex]\frac{2x}{(x+3)(x-1)}[/tex] + [tex]\frac{(x+3)(x+3)}{(x+3)(x-1)}[/tex]
Add the numerators leaving the common denominator ( LCD = (x + 3)(x - 1) )
= [tex]\frac{2x+(x+3)(x+3)}{(x+3)(x-1)}[/tex] ← expand factors on numerator using FOIL
= [tex]\frac{2x+x^2+6x+9}{(x+3)(x-1)}[/tex]
= [tex]\frac{x^2+8x+9}{(x+3)(x-1)}[/tex]
