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Find four consecutive even integers such that the sum of the first three exceeds the fourth by 8.

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Answer:

4, 6, 8, 10

Explanation:

consecutive even integers have a difference of 2 between them.

let the 4 consecutive even integers be

n , n + 2 , n + 4 , n + 6 , then

n + n + 2 + n + 4 = n + 6 + 8

3n + 6 = n + 14 ( subtract n from both sides )

2n + 6 = 14 ( subtract 6 from both sides )

2n = 8 ( divide both sides by 2 )

n = 4

n + 2 = 4 + 2 = 6

n + 4 = 4 + 4 = 8

n + 6 = 4 + 6 = 10

the 4 integers are 4 , 6 , 8 , 10

User Elthrasher
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