(9r2+3r+6)+(6r2+2r)
would the exponents stay two or would they become for, and could you explain to me why?

The terms 9r^2 and 6r^2 are like terms. We add the numbers to the left of the "r^2" to get 9+6 = 15.

So 9r^2+6r^2 becomes 15r^2

The r^2 stays the same. We do not say something like r^2+r^2 = r^4. You might be thinking of r^2 * r^2 = r^(2+2) = r^4

When it comes to adding like terms, we keep the variables the same.

The other pair of like terms are 3r and 2r, so they add to 3r+2r = 5r

Overall,

(9r^2+3r+6)+(6r^2+2r) = (9r^2+6r^2)+(3r+2r)+6

(9r^2+3r+6)+(6r^2+2r) = 15r^2+5r+6

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x