Question

3x3 equations need to find x, y, and z

Answer 1

hmmm usually we'd end up using two pairs of equations to get a resultant equation and then squeeze the system of three into a system of two, but in this case we can do away with only one resultant equation.

So since the 1st and 3rd equations both have "x" and "z", let's use those and eliminate "x" by multiplying the 1st one by "-2".

`\begin{array}{llll} x-2z&=&-5\\ y-3z&=&-3\\ 2x-z&=&-4 \end{array}\implies \begin{array}{llll} x+0y-2z&=&-5\\ 0x+y-3z&=&-3\\ 2x+0y-z&=&-4 \end{array} \\\\\\ \begin{array}{llll} \text{\LARGE -2}( x+0y-2z&=&-5)\\ 2x+0y-z&=&-4 \end{array}\implies \begin{array}{llll} -2x+0y+4z&=&10\\ ~~ 2x+0y-z&=&-4\\\cline{1-3} ~\hfill~3z&=&6 \end{array} \\\\\\ 3z=6\implies z=\cfrac{6}{3}\implies \boxed{z=2} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{let's do substitution on the 1st equation}}{x-2z=-5\implies x-2(2)=-5} \implies x+4=-5\implies \boxed{x=-1} \\\\\\ \stackrel{\textit{now let's do substitution on the 2nd equation}}{y-3z=-3\implies y-3(2)=-3} y-6=-3\implies \boxed{y=3}`