SOLUTION: Find two consecutive odd integers such that seven times the first is eight more than 5 times the second. my answer ; 7x = [5(x+2)] + 8

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Question 920664: Find two consecutive odd integers such that seven times the first is eight more than 5 times the second.
my answer ; 7x = [5(x+2)] + 8
Found 3 solutions by richard1234, josgarithmetic, MathTherapy:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Your on the right track...now solve for x.

Answer by josgarithmetic(39623) About Me  (Show Source):
You can put this solution on YOUR website!
Odd consecutive integers, 2n+1, 2n+3 for any Natural Number n.

highlight_green%287%282n%2B1%29=8%2B5%282n%2B3%29%29

The trouble with your equation compared to this one is that you used variable expressions that
would not ensure to be odd numbers. You can now continue with the equation and solve for n
and then evaluate 2n+1 and 2n+3.

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

Find two consecutive odd integers such that seven times the first is eight more than 5 times the second.
my answer ; 7x = [5(x+2)] + 8


Good work!!



You don't need the brackets. The parentheses are enough. Thus, we have:

7x = 5(x + 2) + 8

Distribute the 5, then combine like-terms, and determine the value of x: the 1st, or 

smaller odd integer. Then add 2 to get the larger odd integer. 



You can do it!!