SOLUTION: Find four consecutive even integers such that two times the sum of the first and second is 40 more than three times the fourth explain

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Find four consecutive even integers such that two times the sum of the first and second is 40 more than three times the fourth explain       Log On


   

Question 1053483: Find four consecutive even integers such that two times the sum of the first and second is 40 more than three times the fourth explain
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let n , n%2B2 , n%2B4 and n%2B6 be the four consecutive even integers.
So,
n%2B%28n%2B2%29= the sum of the first and second integers,
2%28n%2B%28n%2B2%29%29= two times the sum of the first and second integers,
3%28n%2B6%29= three times the fourth integer, and
3%28n%2B6%29%2B40= 40 more than three times the fourth integer.
The problem says that
2%28n%2B%28n%2B2%29%29=3%28n%2B6%29%2B40 ,
and that is the equation we have to solve.

2%28n%2B%28n%2B2%29%29=3%28n%2B6%29%2B40
2%282n%2B2%29=3n%2B18%2B40
4n%2B4=3n%2B58
4n-3n=58-4
highlight%28n=54%29
So the four consecutive even integers are
highlight%2854%29 , highlight%2856%29 , highlight%2858%29 and highlight%2860%29 .

Verification:
The sum of the first and second integers is 54%2B56=110 .
Two times the sum of the first and second integers is 2%2A110=220 .
Three times the fourth integer is 3%2A60=180 .
Forty (40) more than three times the fourth integer is 180%2B40=220 .