find 3 consecutive even integers such that the sum of the smaller 2 is 6 less than twice the largest one
1 answer:
n, n + 2, n + 4 - three consecutive even integers
n + 2 = 2(n + 4) - 6 |use distributive property
n + 2 = (2)(n) + (2)(4) - 6
n + 2 = 2n + 8 - 6
n + 2 = 2n + 2 |subtract 2 from both sides
n = 2n |subtract 2n from both sides
-n = 0
n = 0
n + 2 = 0 + 2 = 2
n + 4 = 0 + 4 = 4
Answer: 0, 2, 4.
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