SOLUTION: This one is tough, 4x=3(4-y),3y=4(2-x) I rewrote the equations to be:4x+3y=12 & 4x-3y= -8 both are systems. The chapter is solving systems of equations by addition. So I mulitp

Question 31320This question is from textbook Beginning and Intermediate Algebra
: This one is tough, 4x=3(4-y),3y=4(2-x)
I rewrote the equations to be:4x+3y=12 & 4x-3y= -8 both are systems. The chapter is solving systems of equations by addition.
So I mulitplied 4(3)+3(3)=12(3) to get 12x+9y=36
Next eq. 4(3)-3(3)= -8 to get 12x-9y= -24
I tried to add and got x=0.5 and I tried to put 0.5 back into the other eq. but something seems wrong. Did I go wrong in my math? Or is this an inconsistent equation?
Thank you, Liz This question is from textbook Beginning and Intermediate Algebra

Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Here's another approach that will lead to the same conclusion that you have reached.
Solve by addition:
1)
2)
First, simplify both equations.
1) Solve this one in terms of y.
2)
1)
2) Multiply this one by -1 then add the two equations.
1)
2)
--------------------
3) NEVER!
The solution is the null or empty set.
The lines are parallel (the slopes are equal) thus there is no intersection, hence, no solution.
Here's the graph:

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