## Question

###### Solve the system of linear equations. $$ \begin{array}{r} 2 x+3 y=6 \\ -x-y=-3 \end{array} $$ Solution: Set up the determinants. $$ D=\left|\begin{array}{rr} 2 & 3 \\ -1 & -1 \end{array}\right|, D_{x}=\left|\begin{array}{rr} 2 & 6 \\ -1 & -3 \end{array}\right|, \text { and } D_{y}=\left|\begin{array}{rr} 6 & 3 \\ -3 & -1 \end{array}\right| $$ Evaluate the determinants. $$ \begin{array}{c} D=1, D_{x}=0, \text { and } D_{y}=3 \\ x=\frac{D_{x}}{D}=\frac{0}{1}=0 \text { and }

Solve the system of linear equations. $$ \begin{array}{r} 2 x+3 y=6 \\ -x-y=-3 \end{array} $$ Solution: Set up the determinants. $$ D=\left|\begin{array}{rr} 2 & 3 \\ -1 & -1 \end{array}\right|, D_{x}=\left|\begin{array}{rr} 2 & 6 \\ -1 & -3 \end{array}\right|, \text { and } D_{y}=\left|\begin{array}{rr} 6 & 3 \\ -3 & -1 \end{array}\right| $$ Evaluate the determinants. $$ \begin{array}{c} D=1, D_{x}=0, \text { and } D_{y}=3 \\ x=\frac{D_{x}}{D}=\frac{0}{1}=0 \text { and } y=\frac{D_{y}}{D}=\frac{3}{1}=3 \end{array} $$ Solve for $x$ and $y .$ $x=0, y=3$ is incorrect. What mistake was made?

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