two positive numbers have a difference of 8 and a product of 33. what are these numbers?
a. 4 and 12
b. 3 and 12
c. 4 and 11
d. 3 and 11

knowing that 33 can only be the product of 1 and 33 or 11 and 3, the answer is 11 and 3 since 11-3 = 8 
but from an algebraic stand point here is how you solve it 
make x be one number and y be the other. You know 
x-y = 8 and 
x * y = 33 
solve either equation for x or y, I will choose the first one and solve for x 
x - y = 8 
x = y+8 
now substitute that value of x in the second equation 
x* y = 33 
(y+8) * y = 33 
y^2 + 8y = 33 
make the equation equal to 0 
y^2 + 8y - 33 = 0 
now factor the equation 
(y+11) (y-3) = 0 
your 2 solutions are 3 and -11, since the question says there are positive numbers, the solution for y is 3 
plug the y value (3) into either equation above and solve for x 
x - 3 = 8 
x = 11