Solve for \(x\) and \(y\)

\begin{align}2x+y & =13 \quad \quad \enclose{circle}{\color{black}{1}}\\ 3x-2y & = 2 \; \quad \quad \enclose{circle}{\color{black}{2}} \end{align}

\begin{align} To \; equate \; coefficients\; of \; y\\ multiply \: eqn \; \enclose{circle}{\color{black}{1}} \; by \; 2\\ 2(2x+y) & = 2(13)\\ 4x+2y & = 26 \quad \;\enclose{circle}{\color{black}{3}}\\ Add \; eqn\; \enclose{circle}{\color{black}{2}} \; to \; eqn \; \enclose{circle}{\color{black}{3}}\\ to \; eliminate \; y \\ (4x+2y)+(3x-2y) & = 26+2\\ 7x & = 28\\ x & = 4\\ By \;substitution \; into\; eqn\; \enclose{circle}{\color{black}{1}}\\ 2(4)+y & = 13\\ y & =5 \end{align}

\begin{align}8y-2x & =28 \quad \quad \enclose{circle}{\color{black}{1}}\\ 6x+7y & = 9 \quad \quad \enclose{circle}{\color{black}{2}} \end{align}

\begin{align} To \; equate \; coefficients\; of \; x\\ multiply \: eqn \; \enclose{circle}{\color{black}{1}} \; by \; 3\\ 3(8y-2x) & = 3(28)\\ 24y-6x & = 84 \quad \enclose{circle}{\color{black}{3}}\\ Add \; eqn\; \enclose{circle}{\color{black}{2}} \; to \; eqn \; \enclose{circle}{\color{black}{3}}\\ to \; eliminate \; x \\ (6x+7y)+(24y-6x) & = 9+84\\ 31y & =93\\ y & =3\\ By \;substitution \; into\; eqn\; \enclose{circle}{\color{black}{1}}\\ 8(3)-2x & = 28\\ -2x & =4\\ x & =-2 \end{align}

\begin{align}5x-2y & =11 \quad \quad \enclose{circle}{\color{black}{1}}\\ 3y+15x & = 6 \quad \quad \enclose{circle}{\color{black}{2}} \end{align}

\begin{align} To \; equate \; coefficients\; of \; x\\ multiply \: eqn \; \enclose{circle}{\color{black}{1}} \; by \; 3\\ 3(5x-2y) & =33\\ 15x-6y & =33 \quad \enclose{circle}{\color{black}{3}}\\ Subtract \; eqn\; \enclose{circle}{\color{black}{2}} \; from \; eqn \; \enclose{circle}{\color{black}{3}}\\ to \; eliminate \; x \\ (15x-6y)-(3y+15x) & = 33-6\\ -9y & = 27\\ y & = -3\\ By \;substitution \; into\; eqn\; \enclose{circle}{\color{black}{1}}\\ 5x-2(-3) & = 11\\ 5x & =5\\ x & =1 \end{align}

\begin{align}-12x+10y & =-28 \quad \quad \enclose{circle}{\color{black}{1}}\\ 3y-2x & = -2 \quad \quad \enclose{circle}{\color{black}{2}} \end{align}

\begin{align} To \; equate \; coefficients\; of \; x\\ multiply \: eqn \; \enclose{circle}{\color{black}{2}} \; by \; 6\\ 6(3y-2x) & = 6(-2)\\ 18y-12x & =-12 \quad \enclose{circle}{\color{black}{3}}\\ Subtract \; eqn \; \enclose{circle}{\color{black}{3}}\; from \; eqn\; \enclose{circle}{\color{black}{1}}\\ to \;eliminate\; x \\ (-12x+10y)-(18y-12x) & = -28-(-12)\\ -8y & = -16\\ y & = 2\\ By \;substitution \; into\; eqn\; \enclose{circle}{\color{black}{1}}\\ -12x+10(2) & = -28\\ -12x= & =-48\\ x & =4 \end{align}