SOLUTION: 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

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!!

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