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Question 17282: Hello! I am trying to solve the system by substitution, this is what I have so far:
5x - 2y = -5
y - 5x = 3
-2y = -5x - 5
y = 5x + 3
-y = 2
5(2) - 2y = -5
10 - 2y = -5
2y = -10 - 5
2y = -5
If I ended this way it would give me a fraction, can a fraction be used with this? Your help is greatly appreciated! :)
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! I think I see an error.
You wrote:
5x - 2y = -5
y - 5x = 3
-2y = -5x - 5
y = 5x + 3
-y = 2
The last line above should be -y = -2, so y = 2.
Then in your next line, I think you made another mistake by substituting the value of y in place of the x. That's a common error!
Now I think it should be this way to finish:
5x- 2y = -5
5x - 2(2) = -5
5x - 4 = -5
5x = -1
x = -1/5
Check: y -5x= 3
2 -5(-1/5)= 3
2+1= 3
It checks!
In answer to your question, YES a fraction CAN be use, and in this case it MUST be used.
R^2 at SCC
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