Question 307440: This one has me stumped: "If 6 is subtracted from the third of three consecutive odd integers and the result is multiplied by 2, the answer is 23 less than the sum of the first and twice the sum of the second of the integers. Find the integers."
My thought process:
1st Int. x
2nd int. x+1
3rd int. x+3
(x-23)+ 2(x+1)+ x+3-6= 2x
or
x + 2(x+1)+ x+3-6= 2x-23
Nothing seems to pan out with this and a few other runs.
Found 2 solutions by mananth, rapaljer:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! This one has me stumped: "If 6 is subtracted from the third of three consecutive odd integers and the result is multiplied by 2, the answer is 23 less than the sum of the first and twice the sum of the second of the integers. Find the integers."
Let the integers be
x, x+2, x+4 ............like 3 , 3+2 , 3+4
2*(x+4 -6)= x + 2(x+2)-23
2*(x-2) = x+2x+4-23
2x-4 = 3x-19
3x-2x= 19-4
x= 15
The numbers are 15, 17, 19
Answer by rapaljer(4671)  (Show Source):
You can put this solution on YOUR website! I'm not quite sure about the wording of the last part of this, but I can tell you that for three consecutive ODD integers, you should use
x= first
x+2= second
x+4= third
The equation will be
2(x+4-6)= x+2(x+2) - 23
See how that works for you.
Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
P.S. If you want to check the wording of the last part and have me finish this for you, then send an Email to Email address at rapaljer@seminolestate.edu. For some reason, the "Thank You" notes apparently are going into the spam filter tonight.
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