SOLUTION: Find 3 consecutive odd integers such that twice the first plus 3 times the third is 19 more than 4 times the second.

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Question 889245: Find 3 consecutive odd integers such that twice the first plus 3 times the third is 19 more than 4 times the second.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
n is any integer.

2n+1, 2n+3, 2n+5 the odd consecutive integers.

2%282n%2B1%29%2B3%282n%2B5%29=19%2B4%282n%2B3%29---this is the description transcribed into symbolism. Solve for n and then use it to compute the requested numbers.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

Find 3 consecutive odd integers such that twice the first plus 3 times the third is 19 more than 4 times the second.


Let 1st of 3 integers be F

Then 2nd and 3rd are: F + 2, and F + 4, respectively

Thus, 2F + 3(F + 4) = 4(F + 2) + 19 

Solve for F: the 1st of the 2 integers, then find 2nd and 3rd.

You can do the check!! 



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