SOLUTION: Find three consecutive even integers such that twice the first minus three times the second plus five times the third is 54.

Question 418432: Find three consecutive even integers such that twice the first minus three times the second plus five times the third is 54.
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
"Find three consecutive even integers such that twice the first minus three times the second plus five times the third is 54."

x = 1st consecutive even integer
x + 2 = 2nd consecutive even integer
x + 4 = 3rd consecutive even integer

2x = "twice the first"
3(x + 2) = "three times the second"
5(x + 4) = "five times the third"

2x - 3(x + 2) + 5(x + 4) = 54
2x - 3x - 6 + 5x + 20 = 54 {used distributive property}
4x + 14 = 54 {combined like terms}
4x = 40 {subtracted 14 from both sides}
x = 10 {divided both sides by 4}
x + 2 = 12 {substituted 10, in for x, into x + 2}
x + 4 = 14 {substituted 10, in for x, into x + 4}

10, 12,and 14 are the three consecutive even integers
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