SOLUTION: 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 =

Algebra ->  Linear-equations -> SOLUTION: 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 =      Log On


   

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) About Me  (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