Question

If 4x^2y^2 + 8x^3y^5/2xy^2 is completely simplified to 2x^ay^b + 4x^cy^d where a,b,c and d represent integer exponents, what is the value of c?

Answer 1

$\frac{4x^2y^2 + 8x^3y^5}{2xy^2}$
remember that (a+b)/c=(a/c)+(b/c)
so
$\frac{4x^2y^2 + 8x^3y^5}{2xy^2}$=$\frac{4x^2y^2}{2xy^2}+ \frac{8x^3y^5}{2xy^2}$

also remember
$\frac{x^m}{x^n}=x^{m-n}$
$\frac{4x^2y^2}{2xy^2}+ \frac{8x^3y^5}{2xy^2}$=
$\frac{2x^2y^2}{xy^2}+ \frac{4x^3y^5}{xy^2}$=
$2x^1y^0+ 4x^2y^3$
a=1
b=0
c=2
d=3