# How do you solve the following linear system: # 3x – 9y = 0, 3x – y = 4 #?

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To solve the linear system:

[ \begin{cases} 3x - 9y = 0 \ 3x - y = 4 \end{cases} ]

First, solve one of the equations for one variable, then substitute this expression into the other equation.

From the second equation, isolate ( x ):

[ 3x - y = 4 \ 3x = y + 4 \ x = \frac{y + 4}{3} ]

Substitute ( x = \frac{y + 4}{3} ) into the first equation:

[ 3\left(\frac{y + 4}{3}\right) - 9y = 0 \ y + 4 - 9y = 0 \ -8y + 4 = 0 \ -8y = -4 \ y = \frac{-4}{-8} \ y = \frac{1}{2} ]

Now that we have found ( y = \frac{1}{2} ), substitute this value into one of the original equations to find ( x ):

[ 3x - 9\left(\frac{1}{2}\right) = 0 \ 3x - \frac{9}{2} = 0 \ 3x = \frac{9}{2} \ x = \frac{9}{6} \ x = \frac{3}{2} ]

So, the solution to the linear system is ( x = \frac{3}{2} ) and ( y = \frac{1}{2} ).

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