Two integers sum to 20, five times the smaller is two more than twice the larger

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The sum of two intergers is twenty. Five times the smaller interger is two more than twice the larger integer. Find the integers.

I'm somehow lost on the problem

I tried setting the problem up to solve for the smaller varrible. Making it x & the larger 20-x. Anyway, I came up with 5x + 2 = 40 - 2x, then 7x = 38... Somewhere I really got screwed up. I know the answers are 6 & 14 but I can't seem to figure out how to setup the problem.
 
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It says "Five times the smaller integer is two more than twice the larger integer"

You wrote that as 5x+2=2(20-x), which is saying "five times the smaller integer plus two is equal to twice the larger integer"

Do you see where you made your mistake now?
 
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Note that if x= 8 y= 6, then x is "two more than" y but x+ 2= 10, not 6. You are "going the wrong way". Instead of 5x+ 2, what should you have?
 
Yes, I got it now. It was late & I miss read the problem. It was 5x = 40-2x+2 which gave me 7x = 42 & the two integers are 6 & 14. Thanks. :approve: