College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 8 - Section 8.3 - Systems of Linear Equations: Determinants - 8.3 Assess Your Understanding - Page 582: 18

Answer

$(-1,2)$

Work Step by Step

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 $D\displaystyle \neq 0,\qquad x=\frac{D_{x}}{D}\quad y=\frac{D_{y}}{D}$ --- $\left\{\begin{array}{l}{x+3 y=5}\\{2x-3y=-8}\end{array}\right. \Rightarrow\left[\begin{array}{ll} a & b\\ c & d \end{array}\right]=\left[\begin{array}{ll} 1 & 3\\ 2 & -3 \end{array}\right],\quad \left[\begin{array}{l} s\\ t \end{array}\right]=\left[\begin{array}{l} 5\\ -8 \end{array}\right]$ $\begin{array}{lllll} D & ... & D_{x} & ... & D_{y}\\ \left|\begin{array}{ll} 1 & 3\\ 2 & -3 \end{array}\right|= & & \left|\begin{array}{ll} 5 & 3\\ -8 & -3 \end{array}\right|= & & \left|\begin{array}{ll} 1 & 5\\ 2 & -8 \end{array}\right|=\\ =-3-6 & & =-15+24 & & =8-10\\ =-9\neq 0 & & =9 & & =-18\\ & & & & \end{array}$ $x=\displaystyle \frac{D_{x}}{D}=\frac{9}{-9}=-1$ $y=\displaystyle \frac{D_{y}}{D}=\frac{-18}{-9}=2$ Solution: $(x,y)=(-1,2)$
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