Find three consecutive even integers such that the sum of twice the first and three times the third is fourteen more than four times the second.

3 consecutive even integers can be represented by 2n, 2n+2 and 2n-2. 2(2n-2)+3(2n+2)=8n+14

4n-4+6n+6=8n+14

2n=14+4-6=12, so n=6

The integers are 10, 12 and 14.

CHECK: Twice the first: 20

plus 3 times the third: 20+42=62

equals 4 times the second plus 14: 48+14=62 Correct!