Question 224375: I need help with systems of equations....Find all the solutions of the follow systems of equations if possible. If a system has no solution, explain why it does not.
(A) 3X-Y=3
3X+Y=15
(B) 2X-Y=1
2Y-4X=3
(C) 2X-Y=1
2Y-4X=(-2)
Please help me understand these, Im lost!!! Thank you
Answer by LtAurora(115)   (Show Source):
You can put this solution on YOUR website! (A) 
There are two ways to solve these types of equations, through substitution method, or through the addition methos. This one lends itself nicely to addition since the  terms are opposite signs.
+
=
This is obtained from adding down.  ,  ,  .
So, to solve for  , we need to divide both sides by 6, which yields  .
We still need to obtain a value for , so we can plug the value into either of our two initial given equations.
I've chosen the second one.
Combine like terms.
Isolate the Y to yield an answer of:
 .
Thus, the answer to this system is,  , .
(B) 2X-Y=1
2Y-4X=3
I'm going to solve this equation with the substitution method, different from the one above.
The first equation needs to be solved for a variable, since the is by itself, we can isolate it fairly easily.
First, subtract the  to the other side of the equation.
 .
Then, to obtain a positive we can multiply by a -1 to switch the signs.
 .
This equation can be substituted into the second equation for :
Combine like terms:
Here the terms cancel to yield:
 .
Since this is not a true statement, there are no solutions to this system. Graphically these would be two parallel lines since they will never intersect ever.
(C) 2X-Y=1
2Y-4X=(-2)
Since the first equation is the same as the one solved above in part (B):
Plugging this into the second equation yields:
Combine like terms:
Here the terms cancel again leaving:
Since this is a true statement, there are an infinite number of solutions. Graphically this would be the same line drawn on itself twice. Thus the two lines are always touching, since they are the same.
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