SOLUTION: find three consecutive even integers, such that the sum of the second and third number is 20 more than the first.

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Question 517004: find three consecutive even integers, such that the sum of the second and third number is 20 more than the first.
Found 2 solutions by jessica43, JBarnum:
Answer by jessica43(140) About Me  (Show Source):
You can put this solution on YOUR website!
To solve this equation you need to write equations using what you know. First, you know that the numbers are 3 consecutive even integers. So A is your first, B = A + 2 and C = B + 2.
Next, you know that the sum of the second and third numbers is 20 more than the first: B + C = 20 + A.
Now plug the first two equations into the second, for B and C:
B + C = 20 + A
(A + 2) + (B + 2) = 20 + A
A + 2 + B + 2 = 20 + A
A + B + 4 = 20 + A
B + 4 = 20
B = 16
So the three consecutive even numbers are 14, 16, 18.

Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
x, x+2, x+4
%28x%2B2%29%2B%28x%2B4%29=20%2Bx add
2x%2B6=20%2Bx subtract x from both sides
x%2B6=20 subtract 6 from both sides
x=14
14%2B2=16
14%2B4=18
highlight%2814%29, highlight%2816%29, highlight%2818%29
check:
16%2B18=34
34-20=14
correct