SOLUTION: The sum of twice one number and 3 times another is 41. If the second is subtracted from the first, the difference is 8. What are the numbers?

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Question 369560: The sum of twice one number and 3 times another is 41. If the second is subtracted from the first, the difference is 8. What are the numbers?
Answer by acalgebra(30) About Me  (Show Source):
You can put this solution on YOUR website!
Let the first number be x and the second number be y, then:
2x%2B3y=41 and x-y=8

The easiest way to solve this is to multiply the second equation by 3:
3x-3y=24
Now, add the equations:
2x%2B3y%2B3x-3y=41%2B24
Combine like terms:
5x=65
Divide both sides by 5 to simplify:
x=13

Now, substitute 13 for each x in one of the original equations. In this case, the second equation will be easier.
13-y=8
Add y to each side:
13=y%2B8
Subtract 8 from each side:
5=y
y=5

So, x=13 and y=5.