Asked by Gabbi
Solve using substitution.If the system has infinitely many solutions, express your answer in terms of k. If the system is inconsistent, enter INCONSISTENT.
Here is my work:
y=5x-5
5x-y=5
5x-(5x-5)=5
5x-5x+5=5
5=5
y=5x-5
k=5x-5
5x=k+5
x=1/5k+1
x,y = (k/5+1, 5k-5)
Here is my work:
y=5x-5
5x-y=5
5x-(5x-5)=5
5x-5x+5=5
5=5
y=5x-5
k=5x-5
5x=k+5
x=1/5k+1
x,y = (k/5+1, 5k-5)
Answers
Answered by
Steve
Once you get to 5=5 you are done. There are infinitely many solutions.
If you put the two equations to the same format, they are
y = 5x-5
y = 5x-5
They are the same line. Any solution to one of them is also a solution to the other.
If you put the two equations to the same format, they are
y = 5x-5
y = 5x-5
They are the same line. Any solution to one of them is also a solution to the other.
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