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,
Penny