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
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-> 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
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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
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)
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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: