SOLUTION: Tony found four consecutive even integers such that 5 times the sum of the first and third was 9604 greater than twice the sum of the second and the fourth. Find the numbers and ex

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Question 1204967: Tony found four consecutive even integers such that 5 times the sum of the first and third was 9604 greater than twice the sum of the second and the fourth. Find the numbers and express each in standard form.
Found 2 solutions by math_tutor2020, mananth:
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

1st number = x = some even integer
2nd number = x+2 = the even integer just after x
3rd number = x+4 = the even integer just after x+2
4th number = x+6 = the even integer just after x+4

One possible equation to form is
5(x + x+4) = 2(x+2 + x+6) + 9604
Pay attention to the color coding to see where all the expressions fit in.

The left side is of the form 5*(1st + 3rd)
The right side is of the form 2*(2nd + 4th) + 9604

I'll let the student solve that equation for x.

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Tony found four consecutive even integers such that 5 times the sum of the first and third was 9604 greater than twice the sum of the second and the fourth. Find the numbers and express each in standard form.


n-3, n-1,n+1,n+3

5 times the sum of the first and third was 9604 greater than twice the sum of the second and the fourth.
5(n-3+n+1)= 2(n-1+n+3)+9604
5(2n-2)=2(2n+2)+9604
10n-10 = 4n+4+9604
6n = 9618
n= 1603
n-3= 1603-3= 1600
n-1= 1603-1= 1602
n+11603+1=1604
n+3= 1603+3- 1606
CHECK
5*(1600+1604)= 2(1602+1606)+9604 =16020