SOLUTION: find four consecutive integers such that the sum of the second and the fourth is seventeen less than thrice the first.

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Question 74306: find four consecutive integers such that the sum of the second and the fourth is seventeen less than thrice the first.
Found 2 solutions by josmiceli, funmath:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the integers a, a+1, a+2, a+3
%28a%2B1%29+%2B+%28a%2B3%29+=+3a+-+17
2a+%2B+4+=+3a+-+17
a+=+21
The integers are 21,22,23,24
check:
22+%2B+24+=+46
3%2A21+-+17+=+46
63+-+17+=+46
46+=+46
OK

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
find four consecutive integers such that the sum of the second and the fourth is seventeen less than thrice the first.
Let the 1st integer be: x
Then the 2nd integer is: x+1
Then the 3rd integer is: x+1+1=x+2
Then the 4th integer is: x+1+1+1=x+3
Sum means:+
17 less means: -17
thrice means: 3*
Problem to solve:
(x+1)+(x+3)=3(x)-17
x+1+x+3=3x-17
2x+4=3x-17
2x-2x+4=3x-2x-17
4=x-17
4+17=x-17+17
21=x
The first integer is: x=21
the second is: x+1=21+1=22
the third is: x+2=21+2=23
the fourth is: x+3=21+3=24
Sanity check:
If we add the 2nd and 4th, do we get 3 times the 1st minus 17?
22+24=3(21)-17
46=63-17
46=46
We seam to be right.
Happy Calculating!!!