I have an algebra 2 question?

x=56

Answer is 58

~David

i'll explain in my words, the average of 3 three #s must be 1/3 of 168, so 168 divided by 3 = 56.

each of the 3 integers have a distance of 2 between each other, so 54, 56, 54 leaves you with 4, the total distances. add that total distance to the 3rd #, so you get 54,56,58. that's your answer. sorry i'm an out-of-the-box type of thinker

Intergers are whole numbers. e.g. 1,2,3

However since it says even* intergers, its like 2,4,6

since they are evenly consecutive

x=x

y=x+2

z=x+2+2=x+4

so x+(x+2)+(x+4) =168

3x=168-4-2=162

x= 162/3=54

since x is 54, y=x+2=54+2=56 and z=x+4=58

see does it work...

54+56+58=168

Largest integer = 58

54 + 56 + 58 = 168

The largest of the three is 58.

3n+3=168

3n=165

n=165/3

165/3=55<-First Integer

55+1=56<-Second Integer

56+1=57<-Third Integer