



 
Hi Lynn, The variable $y$ only appears in the first two equations so I want to collapse them into one equation. Since $2x + y$ is eight and $y + 3z$ is five it must be that $2x + y \mbox{ minus } y + 3z$ is $8  5 = 3.$ Thus \[(2x + y)  (y + 3z) = 8  5 = 3.\] This becomes \[2x + y  y  3z = 3,\] or \[2x  3z = 3.\] This equation together with the third and fourth equations in your problem form three equations with $y$ eliminated. I hope this helps,
 


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