# math

posted by .

I would let #1 = x.
Then #2 = x+1
#3 = x+2
#4 = x+3
The problem says we want the sum of the first three. That will be
x + (x + 1) + (x + 2) and that should be 60 more than the fourth. The fourth is x + 3; therefore, set them equal and add 60 to the fourth. Solve for x.
Post your work if you get stuck.
x+(x+1)+(x+2) = (x+3)+60

Find four consecutive integers such that the sum of the first three is 60 more than the fourth.
What is the best way to solve this?

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