Question

How to determine solutions to a system of linear equations

Answer 1

$\bf \begin{cases} \boxed{y}=-\frac{1}{3}x+2\\\\ x=2y+6 \end{cases} \\\\\\ \textit{using substitution on the \underline{2nd equation}} \\\\\\ x=2\left( \boxed{-\frac{1}{3}x+2} \right)+6\implies x=-\cfrac{2x}{3}+4+6\implies x=-\cfrac{2x}{3}+10 \\\\\\ x=\cfrac{-2x+30}{3}\implies 3x=-2x+30\implies 5x=30 \\\\\\ x=\cfrac{30}{5}\implies \boxed{x=6}$

and since we know now what "x", let's use that in either equation, say the first equation,

$\bf y=-\cfrac{1}{3}x+2\implies y=-\cfrac{1}{3}(6)+2\implies y=-2+2\implies y=0$

another way to do it is by "elimination", depending on the equations, one way may work quicker than the other, in this case, substitution was the simpler one, because the first equation is already solved for "y", so we just grabbed it and plugged it in the 2nd equation.

Answer 2

There are a few ways.
(1) Substitution 
(2) Elimination
(3) Graphing
(4) Different methods using matrices such as putting an augmented matrix into row echelon form or using Cramer's Rule or using the inverse matrix.