Question 513184
Find three consecutive even integers for which -4 times the sum of the first and the third integers is 192.


Let
1st  Interger = x+2
2nd   Interger = x+4
3rd  Integer = x+6

-4*(1st  Integer + 3rd  Integer) = 192
-4(x+2+x+6)=192
-4(2x+8)=192
-8x-32=192
-8x=192+32
-8x=224
-8x/-8=224/-8
x=-28

1st  Interger = x+2 = -28+2=-26
2nd   Interger = x+4 = -28+4=-24
3rd  Integer = x+6 = -24+6=-22

Check
-4*(1st  Integer + 3rd  Integer) = 192
-4*(-26 + (-22)) = 192
-4*(-26-22) = 192
-4*(-48) = 192
192=192