Answer: No real solution.
Step-by-step explanation:
[tex]\frac{2}{(2x+6)}-\frac{2}{x^2+5x+6}=\frac{3}{(x+3)}[/tex]
First solve [tex]x^2+5x+6[/tex]
[tex]\frac{2}{(2x+6)}-\frac{2}{(x+3)(x+2)}=\frac{3}{(x+3)}[/tex]
[tex]\frac{[2(x+3)(x+2)]-[2(2x+6)]}{(2x+6)(x+3)(x+2)} = \frac{3}{(x+3)}[/tex]
[tex]\frac{[(2x+6)(x+2)]-(4x-12)}{(2x+6)(x+3)(x+2)} = \frac{3}{(x+3)}[/tex]
[tex]\frac{(2x^2+4x+6x+12)-(4x-12)}{(2x+6)(x+3)(x+2)} = \frac{3}{(x+3)}[/tex]
[tex]\frac{2x^2+10x+12-4x+12}{(2x+6)(x+3)(x+2)} = \frac{3}{(x+3)}[/tex]
[tex]\frac{2x^2+6x+24}{(2x+6)(x+3)(x+2)} = \frac{3}{(x+3)}[/tex]
Past this point, we can't solve [tex]2x^2+6x+24[/tex] and obtain real values of x. Therefore, there are no real solutions.
