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 761: 31

Answer

$(x,y)=(\dfrac{4}{5},\dfrac{1}{5})$

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} 3 & -5\\ 15 & 5 \end{array}\right]$ and$\left[\begin{array}{l} p\\q \end{array}\right]=\left[\begin{array}{l} 3\\ 21 \end{array}\right]$ $\begin{array}{cccccc} \triangle =& & \triangle_{1} =& & \triangle_{2} = \\\left|\begin{array}{ll} 3 & -5\\ 15 & 5 \end{array}\right|= & & \left|\begin{array}{ll} 3 & -5\\ 21 & 5 \end{array}\right|= & & \left|\begin{array}{ll} 3 & 3\\ 15 & 21 \end{array}\right|=\\ =15+75 & & =15+105 & & =63-45 \\=90 (\ne 0) & & =120 & &=18 \\ & & & & \end{array}$ So, $x= \dfrac{\triangle_{1}}{\triangle}=\dfrac{120}{90}=\dfrac{4}{3}$ and $y=\dfrac{\triangle_{2}}{\triangle}=\dfrac{18}{90}=\dfrac{1}{5}$ Thus, our solution is: $(x,y)=(\dfrac{4}{3},\dfrac{1}{5})$
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