Question 74306
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!!!