Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 10 - Systems of Equations and Inequalities - Section 10.3 Systems of Linear Equations: Determinants - 10.3 Assess Your Understanding - Page 760: 15

Answer

$(x,y)=(6,2)$

Work Step by Step

Cramer's rule states that $a x+b y=p \\ cx+dy=q$ $\triangle=\left|\begin{array}{ll}{a}&{b}\\{c}&{d}\end{array}\right|, \triangle_{1}=\left|\begin{array}{ll}{p}&{b}\\{q}&{d}\end{array}\right|, \triangle_{2}=\left|\begin{array}{ll}{a}&{p}\\{c}&{q}\end{array}\right|$; $ x=\dfrac{\triangle_1}{\triangle}; y=\dfrac{\triangle_2}{\triangle} (D\displaystyle \neq 0)$ From the given system of equations, we have: $ \left[\begin{array}{ll} a & b\\ c & d \end{array}\right]=\left[\begin{array}{ll} 1 & 1\\ 1 & -1 \end{array}\right],\quad \left[\begin{array}{l} p\\ q \end{array}\right]=\left[\begin{array}{l} 8\\ 4 \end{array}\right]$ $\begin{array}{cccccc} \triangle =& & \triangle_{1} =& & \triangle_{2} = \\ \left|\begin{array}{ll} 1 & 1\\ 1 & -1 \end{array}\right|= & & \left|\begin{array}{ll} 8 & 1\\ 4 & -1 \end{array}\right|= & & \left|\begin{array}{ll} 1 & 8\\ 1 & 4 \end{array}\right|=\\ =-1-1 & & =-8-4 & & =4-8\\ =-2 & & =-12 & & =-4\\ & & & & \end{array}$ So, $x= \dfrac{\triangle_{1}}{\triangle}=\dfrac{-12}{-2}=6$ and $y=\dfrac{\triangle_{2}}{\triangle}=\dfrac{-4}{-2}=2$ Thus, our solution is: $(x,y)=(6,2)$
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