SOLUTION: Find three consecutive odd integers such that -3 times the sum of the first and third is 50 greater than 8 times the opposite of the second.

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Find three consecutive odd integers such that -3 times the sum of the first and third is 50 greater than 8 times the opposite of the second.      Log On


   

Question 1202493: Find three consecutive odd integers such that -3 times the sum of the first and third is 50 greater than 8 times the opposite of the second.
Found 2 solutions by Theo, josgarithmetic:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
x is the first number.
x + 2 is the second number.
x + 4 is the third number.
-3 * (x + x + 4) = 8 * - (x + 2) + 50
simplify to get:
-3 * (2x + 4) = -8 * x - 16 + 50
simplify further to get:
-6x - 12 = -8x + 34
add 8x to both sides of the equation and add 12 to both sides of the equation to get:
2x = 46
solve for x to get:
x = 23
first number is 23
second number is 25
third number is 27
-3 times the sum of the first and third is 50 greater than 8 times the opposite of the second becomes:
-3 * (23 + 27) = 8 * -25 + 50
this becomes -3 * 50 = -200 + 50 which becomes:
-150 = -150, confirming the values of the 3 consecutive odd integers are correct.
your solution is the the 3 consercutive odd integers are 23, 25, 27.

Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
consecutive odd
n, n+2, n+4

description
-3%28n%2Bn%2B4%29=50%2B8%28-1%29%28n%2B2%29
simple algebra from here.


-6n-12=50-8n-16
2n=12%2B50-16
n=6%2B25-8
n=23
.
.
23, 25, 27