Question 1171400

I am having trouble with understanding this question:

The sum of three numbers is 40. The second number is twice the first number. The sum of the first two numbers is 2 more than the third number. What are the three numbers?

So far, I've worked out:
x+y+z=10
y=2x
x+y=z+2
<pre>The setup of the equations is fine, with the exception of the 1st, which has a sum of 40, not 10..
x + y + z = 40 ------ eq (i)
y = 2x ------- eq (ii)
x + y = z + 2_____x + y - z = 2 ------- eq (iii)
2z = 38 ------ Subtracting eq (iii) from eq (i) 
{{{highlight_green(matrix(1,5, z, "=", 38/2, "=", 19))}}}

2x + 2y = 42 ------ Adding eq (iii) & (i) 
  x + y = 21 ------ eq (iv)
 x + 2x = 21 ------ Substituting 2x for y in eq (iv)
     3x = 21
{{{highlight_green(matrix(1,5, x, "=", 21/3, "=", 7))}}}

{{{highlight_green(matrix(1,5, y, "=", 2(7), "=", 14))}}} ----- Substituting 7 for y in eq (ii)</pre>