Anumber doubled and has four added to it. the result is divided by three and then added to the product of the original number and two. the resulting sum is twenty.

X= a number

(2x+4)/3 + (x+2)= 20
Multiply everything by 3 to eliminate the fraction
(3/1)((2x+4)/3) + (3)(x+2)= (3)(20)
2x+4+3x+6= 60
5x+10=60
Subtract 10 from both sides
5x=50
divide both sides by 5
x=10

Check:
Substitute answer in original equation
(2x+4)/3 + (x+2)= 20
(2(10)+4)/3+(10+2)= 20
(20+4)/3+12= 20
24/3+12= 20
8+12= 20
20=20

Hope this helps! :)