Question 73920
Let n=1st number, o=2nd number, p=3rd number, q=4th number.
Since they are consecutive, the 2nd number is 1 more than the 1st number, the 3rd is 1 more than the 2nd, etc. So it looks like this
*[Tex \large o=n+1 \\ p=o+1 \\ q=p+1]
*[Tex \large n+o+p=q] And the sum of the first 3 equal the 4th
Solve for o for the 2nd equation
*[Tex \large o=p-1]
Solve for n for the 1st equation
*[Tex \large n=o-1]
Plug these new substitutions into *[Tex \large n+o+p=q] and plug in p+1 into q
*[Tex \large (o-1)+(p-1)+p=(p+1)]
Now plug in p-1 into o
*[Tex \large ((p-1)-1)+(p-1)+p=(p+1)]
Now we have an equation with nothing but p's. We can now solve for p
*[Tex \large 3p-3=p+1]
*[Tex \large 2p=4]
*[Tex \large  p=2]
Use this to solve for the rest of the numbers
*[Tex \large o=p-1]
*[Tex \large o=2-1]
*[Tex \large o=1]
Use this to solve for the rest of the numbers
*[Tex \large n=o-1]
*[Tex \large n=1-1]
*[Tex \large n=0]
So n=0, o=1, and p=2. That means q=3
*[Tex \large n+o+p=q]
*[Tex \large 0+1+2=3]
So the numbers are 0,1,2,3
Hope this makes sense.