Working:

Z=3-2y, given because 2y+z=3.

$\displaystyle 3c+4y+z=7 \rightarrow 3c+4y+3-2y=7 \rightarrow 3c+2y+3=7$so $\displaystyle 3x+2y+3-3=7-3 \rightarrow 3x+2y=4 \therefore y=2-\frac{3}{2}x$

My calculations break down here, where I try to solve the third equation for x...

I can't seem to finish the operation.

Here is the attempted working:

[Math]-5c+3y+8z=37 \rightarrow -5+3(2-\frac{3}{2}x)+8z=-31 \rightarrow -5x+6-\frac{9}{2}x+8z=-31 \rightarrow 9\frac{1}{2}C+6+8z=-31[/tex]so $\displaystyle -9\frac{1}{2}C+6-6+8z=-31-6 \rightarrow -9\frac{1}{2}x+8z=-37$

This is the point at which my calculations fall apart.

Thank you!