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

\begin{align}4x+y & =10 \quad \quad \enclose{circle}{\color{black}{1}}\\ 3x-y & = 11 \quad \quad \enclose{circle}{\color{black}{2}} \end{align}

\begin{align} Add \; eqn \; \enclose{circle}{\color{black}{1}}\; to \; eqn\; \enclose{circle}{\color{black}{2}}\\ to \;eliminate\; y \\ (4x+y)+(3x-y) & = 10+11\\ 7x & = 21\\ x & = 3\\ Then \; by \;substitution \\ into\; eqn\; \enclose{circle}{\color{black}{1}}\\ 4(3)+y & = 10\\ y & =-2 \end{align}

\begin{align}4y+3x & =20 \quad \quad \enclose{circle}{\color{black}{1}}\\ -3x+2y & = -8 \quad \quad \enclose{circle}{\color{black}{2}} \end{align}

\begin{align} Add \; eqn \; \enclose{circle}{\color{black}{1}}\; to \; eqn\; \enclose{circle}{\color{black}{2}}\\ to \;eliminate\; x \\ (4y+3x)+(-3x+2y) & = 20-8\\ 6y & = 12\\ y & = 2\\ Then \; by \;substitution \\ into\; eqn\; \enclose{circle}{\color{black}{1}}\\ 4(2)+3x & = 20\\ 3x & =12\\ x & =4 \end{align}

\begin{align}6x+4y & =18 \quad \quad \enclose{circle}{\color{black}{1}}\\ 2x+4y & = 14 \quad \quad \enclose{circle}{\color{black}{2}} \end{align}

\begin{align} Subtract \; eqn \;\enclose{circle}{\color{black}{2}}\; from \; eqn\; \enclose{circle}{\color{black}{1}}\\ to \;eliminate\; y \\ (6x+4y)-(2x+4y) & = 18-14\\ 4x & = 4\\ x & = 1\\ Then \; by \;substitution \\ into\; eqn\; \enclose{circle}{\color{black}{1}}\\ 6(1)+4y & = 18\\ 4y & =12\\ y & =3 \end{align}

\begin{align}-5x+2y & =2 \quad \quad \enclose{circle}{\color{black}{1}}\\ y-5x & = -4 \quad \quad \enclose{circle}{\color{black}{2}} \end{align}

\begin{align} Subtract \; eqn \; \enclose{circle}{\color{black}{2}}\; from \; eqn\; \enclose{circle}{\color{black}{1}}\\ to \;eliminate\; x \\ (-5x+2y)-(y-5x) & = 2-(-4)\\ y & = 6\\ Then \; by \;substitution \\ into\; eqn\; \enclose{circle}{\color{black}{1}}\\ -5x+2(6) & = 2\\ -5x= & =-10\\ x & =2 \end{align}