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: 25

Answer

$(\displaystyle \frac{1}{2},\frac{3}{4})$

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 $D\displaystyle \neq 0,\qquad x=\frac{D_{x}}{D}\quad y=\frac{D_{y}}{D}$ --- $\left\{\begin{array}{l}{2 x-4 y=-2}\\{3x+2y=3}\end{array}\right.\Rightarrow\left[\begin{array}{ll} a & b\\ c & d \end{array}\right]=\left[\begin{array}{ll} 2 & -4\\ 3 & 2 \end{array}\right],\quad \left[\begin{array}{l} s\\ t \end{array}\right]=\left[\begin{array}{l} -2\\ 3 \end{array}\right]$ $\begin{array}{lllll} D=\left|\begin{array}{ll} 2 & -4\\ 3 & 2 \end{array}\right|= & & D_{x}=\left|\begin{array}{ll} -2 & -4\\ 3 & 2 \end{array}\right|= & & D_{y}=\left|\begin{array}{ll} 2 & -2\\ 3 & 3 \end{array}\right|=\\ =4+12 & & =-4+12 & & =6+6\\ =16\neq 0 & & =8 & & =12\\ & & & & \end{array}$ $x=\displaystyle \frac{D_{x}}{D}=\frac{8}{16}=\frac{1}{2}$ $y=\displaystyle \frac{D_{y}}{D}=\frac{12}{16}=\frac{3}{4}$ Solution: $(x,y)=(\displaystyle \frac{1}{2},\frac{3}{4})$
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