Question 1065096
Find three consecutive even integers such that four times the third integer is four more than three times the first integer plus two times the second integer. 
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Three consecutive even integers: n, (n+2), (n+4)
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write an equation for exactly what it says
4(n+4) = 3n + 4 + 2(n+2)
Distribute, combine like terms
4n + 16 = 3n + 4 + 2n + 4
4n + 16 = 5n + 8
16 - 8 = 5n - 4n
8 = n
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8, 10, 12 are the integers
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Check solution in the statement
" four times the third integer is four more than three times the first integer plus two times the second integer."
4(12) = 3(8) + 4 + 2(10)
48 = 24 + 4 + 20