Answer:
The three numbers are 32,33, and 34.
Step-by-step explanation:
Let the three consecutive integers be [tex]n-1,n, \text{ and } n+1[/tex] for any integer [tex]n[/tex].
Integers are numbers in the set: [tex]\{...,-3,-2,-1,0,1,2,3,...\}[/tex].
We are given that '3 times greatest of these numbers exceeds 2 times the smallest of these numbers by 38'.
This means as an equation we have:
[tex]3(n+1)=38+2(n-1)[/tex].
Let's solve the above equation.
Distribute:
[tex]3n+3=38+2n-2[/tex]
Combine like terms on the sides:
[tex]3n+3=2n+36[/tex]
Subtract [tex]2n[/tex] on both sides:
[tex]n+3=36[/tex]
Subtract 3 on both sides:
[tex]n=33[/tex]
Now let's find [tex]n-1[/tex] and [tex]n+1[/tex] given that [tex]n=33[/tex]:
[tex]n-1=33-1=32[/tex]
[tex]n+1=33+1=34[/tex]
So the three numbers are 32,33, and 34.
Let's check.
3(greatest)=3(34)=102
2(least)=2(32)=64
Now let's see if 102 is 38 more than 64.
Since 38+64=102, it is.
