SOLUTION: 1.find four consecutive integers such that twice the smallest is 12 more than the largest 2.find three consecutive integers such that the sum of the first two integers is 24 more

Algebra ->  Real-numbers -> SOLUTION: 1.find four consecutive integers such that twice the smallest is 12 more than the largest 2.find three consecutive integers such that the sum of the first two integers is 24 more       Log On


   



Question 169688: 1.find four consecutive integers such that twice the smallest is 12 more than the largest
2.find three consecutive integers such that the sum of the first two integers is 24 more than the third integer.
3.find three consecutive odd integers such that the sum of the first and the second is 27 less than 3 times the third
4.find three consecutive even integers such that the sum of the smallest and twice the second is 20 more than the third

Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
I will show you 2 problems 1 and 4 and then you should get the pattern of how its done, remembering that consecutive intergers with a given variable a is a,a+1,a+2 etc., whereas consecutive odd or even intergers would be written as a,a+2,a+4 etc. with that background.
1.) lets call our 4 consecutive integers a,a+1,a+2,a+3
:
2 times the smallest which is a=2a....12 more than largest(a+3) is (a+3)+12
:
2a=(a+3)+12 subtract a from each side
highlight%28a=15%29 so our integers are 15,16,17,18
:
4.) Lets call our 3 consecutive even integers a, a+2,a+4
sum of the smallest(a) plus twice the second(2(a+2)) equals 20 more than the 3rd((a+4)+20)
:
2a+2(a+2)=(a+4)+20
3a+4=a+24---->subtract a and -4 from both sides
2a=20
highlight%28a=10%29 so our even integers are 10,12,14