SOLUTION: Find the smallest of four consecutive even integers if twice the sum of the second and thrird ingers is equal to the sum of the first and the fourth integers increase by ten.

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Question 201056This question is from textbook
: Find the smallest of four consecutive even integers if twice the sum of the second and thrird ingers is equal to the sum of the first and the fourth integers increase by ten. This question is from textbook

Found 2 solutions by josmiceli, RAY100:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the consecutive integers from lowest to highest be
n, n%2B1, n%2B2, n%2B3
given:
2%2A%28n+%2B+1+%2B+n+%2B+2%29+=+n+%2B+n+%2B+3+%2B+10
2n+%2B+2+%2B+2n+%2B+4+=+2n+%2B+13
4n+%2B+6+=+2n+%2B+13
2n+=+7
n+=+7%2F2
Can't be right, since it's not an integer.
Can you see an error? I can't.

Answer by RAY100(1637) About Me  (Show Source):
You can put this solution on YOUR website!
let integers be, x,,,x+2,,,x+4,,,x+6
.
2[(x+2) +(x+4)] = [(x) + (x+6)] +10
.
2[2x +6] = [2x +6] +10
.
4x +12 = 2x +16
.
2x = 4
.
x=2,,,{2,,,,4,,,,6,,,,8,,,,,,,integers}
.
check
.
2(4+6) = (2 + 8) +10
.
20 =10 +10,,,,,,ok