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Solve the system. y = -1/3 x + 2 and x + 3y = 3
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Plug in the first equation into the second one. Then solve for x.
yes, I suggest you use substitution method for fractions. It's much easier. So you have this system: \[\left\{\begin{matrix} \sf \color{red}{y=-\frac{1}{3}x+2} & \\ \sf \color{blue}{y + 3}\sf \color{red}{y} \sf \color{blue}{=3} & \end{matrix}\right.\] plug in the red requation into the blue one where the red y is.
sorry, the first blue \(y\) should be a \(\sf \color{blue}{x}\)
@gabbie96 you can use Matrices to solve this equation.
Why the heck would you want to use matrices tho.
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They are parallel lines look closely
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