Question 627193: Find the largest of three consecutive odd integers, such that 4 times the middle integer is 10 more than the sum of the first and third.
Please help with setting up the equation
Found 2 solutions by jim_thompson5910, MathTherapy:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! First odd integer: 2x+1
Next consecutive odd integer: (2x+1)+2 = 2x+3
Next consecutive odd integer: (2x+3)+2 = 2x+5
So the three consecutive odd integers (in order from smallest to largest) are 2x+1, 2x+3, and 2x+5
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We're given that "4 times the middle integer is 10 more than the sum of the first and third", which translates to
4 times middle = First + Third + 10
4*(2x+3) = (2x+1) + (2x+5) + 10
4*(2x+3) = 2x+1 + 2x+5 + 10
So your task is to solve
4*(2x+3) = 2x+1 + 2x+5 + 10
Then use the solution for x to determine
2x+1, 2x+3, and 2x+5
Let me know if this helps or not. Thanks.
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Answer by MathTherapy(10557)  (Show Source):
You can put this solution on YOUR website!
Find the largest of three consecutive odd integers, such that 4 times the middle integer is 10 more than the sum of the first and third.
Please help with setting up the equation
Let the largest integer be L
Then middle integer = L - 2, and first, or smallest = L - 4
We therefore have: 4(L - 2) = L - 4 + L + 10
4L - 8 = 2L + 6
You should be able to take it from here!!
Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
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