157k views
5 votes
Find four consecutive integers such that twice the 3rd number decreased by the second numberis 8

1 Answer

6 votes

Let's call a, b, c and d he four consecutive integers.

Then, twice the 3rd number decreased by the second number is 8, so:

2c - b = 8

They are consecutive so:

b = a + 1

c = a + 2

d = a + 3

Replacing on the first equation and solving for a, we get:

2c - b = 8

2(a+ 2) - (a + 1) = 8

2a + 4 - a - 1 = 8

a + 3 =

User Gnadelwartz
by
4.5k points