## College Algebra (10th Edition)

$(8,-4)$
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 $D\displaystyle \neq 0,\qquad x=\frac{D_{x}}{D}\quad y=\frac{D_{y}}{D}$ --- $\left\{\begin{array}{r}{3 x=24}\\{x+2y=0}\end{array}\right.\Rightarrow\left[\begin{array}{ll} a & b\\ c & d \end{array}\right]=\left[\begin{array}{ll} 3 & 0\\ 1 & 2 \end{array}\right],\quad \left[\begin{array}{l} s\\ t \end{array}\right]=\left[\begin{array}{l} 24\\ 0 \end{array}\right]$ $\begin{array}{lllll} D & ... & D_{x} & ... & D_{y}\\ \left|\begin{array}{ll} 3 & 0\\ 1 & 2 \end{array}\right|= & & \left|\begin{array}{ll} 24 & 0\\ 0 & 2 \end{array}\right|= & & \left|\begin{array}{ll} 3 & 24\\ 1 & 0 \end{array}\right|=\\ =6-0 & & =48-0 & & =0-24\\ =6\neq 0 & & =48 & & =-24\\ & & & & \end{array}$ $x=\displaystyle \frac{D_{x}}{D}=\frac{48}{6}=8$ $y=\displaystyle \frac{D_{y}}{D}=\frac{-24}{6}=-4$ Solution: $(x,y)=(8,-4)$