Find four consecutive integers such that the sum of the two largest subtracted from three times the sum of the two smallest is 70

18, 19, 20, 21

Step-by-step explanation:

Just like any of these problems, we should start by forming an equation for us to get a reference and plug in. We'll be using x as our variables.

As we have four consecutive integers (and not multiples) we can assume that the integers will be x, x+1, x+2, and x+3.

The sum of the two largest integers we have equals: n+2 + n+3 = 2n+5

and three times the sum of the two smallest = 3(n + n+1) = 6n+3

and the sum of t he two largest subtracted from three times the sum of the two smallest = (6n+3) - (2n+5) = 4n-2

4n-2=70

4n=68

n = 18.

n = 18, n+1 = 19, n+2 = 20, n+3 = 21