## College Algebra (10th Edition)

$(2,-3)$
Cramer's rule $\left\{\begin{array}{l}{a x+b y=s}\\{cx+dy=t}\end{array}\right.$ $D=\left|\begin{array}{ll}{a}&{b}\\{c}&{d}\end{array}\right|,D_{x}=\left|\begin{array}{ll}{s}&{b}\\{t}&{d}\end{array}\right|,D_{y}=\left|\begin{array}{ll}{a}&{s}\\{c}&{t}\end{array}\right|,$ If $D\displaystyle \neq 0,\qquad x=\frac{D_{x}}{D}\quad y=\frac{D_{y}}{D}$ --- \displaystyle \left\{\begin{aligned}\frac{1}{2}x+y&=-2\displaystyle \\x-2y&=8\displaystyle \end{aligned}\right.\Rightarrow\left[\begin{array}{ll} a & b\\ c & d \end{array}\right]=\displaystyle \left[\begin{array}{ll} 1/2 & 1\\ 1 & -2 \end{array}\right],\quad \left[\begin{array}{l} s\\ t \end{array}\right]=\left[\begin{array}{l} -2\\ 8 \end{array}\right] $\begin{array}{lllll} D=\left|\begin{array}{ll} 1/2 & 1\\ 1 & -2 \end{array}\right|= & & D_{x}=\left|\begin{array}{ll} -2 & 1\\ 8 & -2 \end{array}\right|= & & D_{y}=\left|\begin{array}{ll} 1/2 & -2\\ 1 & 8 \end{array}\right|=\\ =-1-1 & & =4-8 & & =4+2\\ =-2\neq 0 & & =-4 & & =6\\ & & & & \end{array}$ $x=\displaystyle \frac{D_{x}}{D}=\frac{-4}{-2}=2$ $y=\displaystyle \frac{D_{y}}{D}=\frac{6}{-2}=-3$ Solution: $(x,y)=(2,-3)$