2 Again, to solve with elimination we’re going to find a way to multiply one of the equations by something so that either the x-coefficients in each equation are equal, or the y-coefficients are equal. I see an easy way to do that:

    \begin{align*}\dfrac{1}{3}x+\dfrac{1}{6}y&=12\\\\ 6\left(\dfrac{1}{3}x+\dfrac{1}{6}y\right)&=6(12)\\\\ 2x+y&=72 \end{align*}

There—now we have a 2x in both equations….

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