## College Algebra (10th Edition)

$(\displaystyle \frac{1}{10},\frac{2}{5})$
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{aligned}2x-3y&=-1\\10x+10y&=5\end{aligned}\right.\Rightarrow\left[\begin{array}{ll} a & b\\ c & d \end{array}\right]=\left[\begin{array}{ll} 2 & -3\\ 10 & 10 \end{array}\right],\quad \left[\begin{array}{l} s\\ t \end{array}\right]=\left[\begin{array}{l} -1\\ 5 \end{array}\right] $\begin{array}{lllll} D=\left|\begin{array}{ll} 2 & -3\\ 10 & 10 \end{array}\right|= & & D_{x}=\left|\begin{array}{ll} -1 & -3\\ 5 & 10 \end{array}\right|= & & D_{y}=\left|\begin{array}{ll} 2 & -1\\ 10 & 5 \end{array}\right|=\\ =20+30 & & =-10+15 & & =10+10\\ =50\neq 0 & & =5 & & =20\\ & & & & \end{array}$ $x=\displaystyle \frac{D_{x}}{D}=\frac{5}{50}=\frac{1}{10}$ $y=\displaystyle \frac{D_{y}}{D}=\frac{20}{50}=\frac{2}{5}$ Solution: $(x,y)=(\displaystyle \frac{1}{10},\frac{2}{5})$