Question 1171400
The sum of three numbers is 40. The second number is twice the first numxber. 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=40
y=2x
x+y=z+2
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Your eqns are correct except for the sum = 10
Sub for y in the 1st and 3rd eqns.
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x + 2x + z = 40 ---> 3x + z = 40
x + 2x = z + 2 ---> 3x = z + 2 --> 3x - z = 2
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3x + z = 40
3x - z = 2
------------------- Add
6x = 42
x = 7
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Solve for y, then solve for z