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: 29

Answer

$(x,y)=(\dfrac{3}{2},1)$

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} 2 & 3\\ 1 & -1 \end{array}\right]$ and$\left[\begin{array}{l} p\\q \end{array}\right]=\left[\begin{array}{l} 6\\ \dfrac{1}{2} \end{array}\right]$ $\begin{array}{cccccc} \triangle =& & \triangle_{1} =& & \triangle_{2} = \\\left|\begin{array}{ll} 2 & 3\\ 1 & -1 \end{array}\right|= & & \left|\begin{array}{ll} 6 & 3\\ \dfrac{1}{2} & -1 \end{array}\right|= & & \left|\begin{array}{ll} 2 & 6\\ 1 & \dfrac{1}{2} \end{array}\right|=\\ =-2-3 & & =-6-\dfrac{3}{2}& & =1-6 \\=-5 (\ne 0) & & =\dfrac{-15}{2} & & =-5\\ & & & & \end{array}$ So, $x= \dfrac{\triangle_{1}}{\triangle}=\dfrac{-15/2}{-5}=\dfrac{3}{2}$ and $y=\dfrac{\triangle_{2}}{\triangle}=\dfrac{-5}{-5}=1$ Thus, our solution is: $(x,y)=(\dfrac{3}{2},1)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.