Algebra II Problem for Released CST

The answer is A. (as it says on the website)
But how do you actually solve this problem? I need help, so when I take the CST (California State Test) tomorrow, I will know how to do it. Thank you.
Here's the link: http://www.cde.ca.gov/ta/tg/sr/documents…

18.
(x+3)/(x+5) + 6/(x^2+3x-10)
=(x+3)/(x+5) + 6/(x+5)(x-2)
=(x+3)(x-2)/(x+5)(x-2) + 6/(x+5)(x-2)
= (x^2+x-6)/(x+5)(x-2) + 6/(x+5)(x-2)
= (x^2+x-6+6)/(x+5)(x-2)
=(x^2+x)/(x+5)(x-2)
=(x^2+x)/(x^2+3x-10)

20.
(20x^-4 / 27^2 )/(8x^-3 / 15y^-5)
=(20/27x^4y^2) / (8y^5 / 15x^3)
=(20/27x^4y^2)*(15x^3 / 8y^5)
= 25 / 18xy^7 => D

21.
4x^2-16 /2-x
=4(x^2-4) / 2-x
=4(x+2)(x-2) /(2-x)
=4(x+2)(x-2) / -(x-2)
= -4(x+2) => D

22.
x^2+4x / x+3 * x^2-9 / x^2+x-12
=x(x+4) /(x+3) * (x+3)(x-3) / (x+4)(x-3)
= x => B

practice more factor will help you.