Answer :
Substitution:
[tex]y=-3x+2 \\ y=2x-3 \\ \\ \hbox{substitute -3x+2 for y in the 2nd equation and solve for x:} \\ -3x+2=2x-3 \\ -3x-2x=-3-2 \\ -5x=-5 \\ x=1 \\ \\ \hbox{substitue 1 for x in one of the equations and solve for y:} \\ y=2 \times 1-3 \\ y=2-3 \\ y=-1 \\ \\ (x,y)=(1,-1)[/tex]
Elimination:
[tex]y=-3x+2 \ \ \ |\times (-1) \\ y=2x-3 \\ \\ -y=3x-2 \\ \underline{y=2x-3} \\ y-y=3x+2x-2-3 \\ 0=5x-5 \\ 5=5x \\ x=1 \\ \\ y=2 \times 1-3=2-3=-1 \\ \\ (x,y)=(1,-1)[/tex]
[tex]y=-3x+2 \\ y=2x-3 \\ \\ \hbox{substitute -3x+2 for y in the 2nd equation and solve for x:} \\ -3x+2=2x-3 \\ -3x-2x=-3-2 \\ -5x=-5 \\ x=1 \\ \\ \hbox{substitue 1 for x in one of the equations and solve for y:} \\ y=2 \times 1-3 \\ y=2-3 \\ y=-1 \\ \\ (x,y)=(1,-1)[/tex]
Elimination:
[tex]y=-3x+2 \ \ \ |\times (-1) \\ y=2x-3 \\ \\ -y=3x-2 \\ \underline{y=2x-3} \\ y-y=3x+2x-2-3 \\ 0=5x-5 \\ 5=5x \\ x=1 \\ \\ y=2 \times 1-3=2-3=-1 \\ \\ (x,y)=(1,-1)[/tex]
[tex]Elimination\ method:\\\\ \left\{\begin{array}{ccc}y=-3x+2&|add\ 3x\ to\ both\ sides\\y=2x-3&|subtract\ "y"\ from\ both\ sides\ and\ add\ 3\ to\ both\ sides\end{array}\right\\\\ \left\{\begin{array}{ccc}3x+y=2\\3=2x-y&|
change\ sides\ of\ the\ equation\end{array}\right\\\\+\underline{\left\{\begin{array}{ccc}3x+y=2\\2x-y=3\end{array}\right}\ \ \ \ |add\ sides\ of\ the\ equations\\. \ \ \ \ \ 5x=5\ \ \ \ \ |divide\ both\ sides\ by\ 5\\.\ \ \ \ \ \ \boxed{x=1}[/tex]
[tex]subtitute\ value\ of\ "x"\ to\ the\ equation\ 3x+y=2:\\\\3(1)+y=2\\3+y=2\ \ \ \ |subtract\ 3\ from\ both\ sides\\\boxed{y=-1}\\\\Answer:\boxed{ \left\{\begin{array}{ccc}x=1\\y=-1\end{array}\right}[/tex]
[tex]subtitute\ value\ of\ "x"\ to\ the\ equation\ 3x+y=2:\\\\3(1)+y=2\\3+y=2\ \ \ \ |subtract\ 3\ from\ both\ sides\\\boxed{y=-1}\\\\Answer:\boxed{ \left\{\begin{array}{ccc}x=1\\y=-1\end{array}\right}[/tex]
