Answer:
B. equivalent
Step-by-step explanation:
Our two equations are:
[tex]y=-\frac{1}{3} x +\frac{2}{3}[/tex]
[tex]2x+6y=4[/tex]
Let's use substitution and plug in the expression [tex]-\frac{1}{3} x +\frac{2}{3}[/tex] for y in the second equation. That way, we're getting rid of a variable so we can solve for the other:
[tex]2x+6y=4[/tex]
[tex]2x+6(-\frac{1}{3} x +\frac{2}{3})=4[/tex]
[tex]2x-2x+4=4[/tex]
[tex]4=4[/tex]
[tex]0=0[/tex]
This means that there are an infinite number of solutions possible for this system - any x value will work. So, the answer is equivalent, because these two equations are essentially the same.
