How do you find four consecutive integers such that the sum of the two largest subtracted from twice the sum of the two smallest is 15?

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Answer 1 · 2015-06-18T23:10:14

Substitute #n#, #n+1#, #n+2# and #n+3# into the description to get an equation in #n# and solve it to find #n=9#.

Hence the integers are: #9#, #10#, #11# and #12#.

Let the four integers be #n#, #n+1#, #n+2# and #n+3#

The sum of the two largest integers is:

#(n+2)+(n+3) = 2n+5#

Twice the sum of the two smallest integers is:

#2(n + (n+1)) = 2(2n+1) = 4n+2#

We are given:

#15 = 2(n + (n+1)) - ((n+2)+(n+3))#

#= (4n+2) - (2n+5) = 2n-3#

Add #3# to both ends to get: #2n=18#

Divide both sides by #2# to get #n=9#

So the four integers are: #9#, #10#, #11#, #12#