Find the three consecutives even integers such that three times the sum of the first and third integers is twenty more than four times the middle number.

Answer 1

3(F+T) = 20+ 4M

3(x-2+x+2) = 20+4x

6x =20+4x

2x=20

x=10

x-2= 8

x+2 = 12

3(8+12) = 20+4(10)

60=60

the numbers are 8, 10 and 12