Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 11 - Systems of Equations and Inequalities - 11.3 Systems of Linear Equations: Determinants - 11.3 Assess Your Understanding - Page 742: 30

Answer

$(x,y) =(2,-3)$

Work Step by Step

The given system of equations is $\left\{\begin{matrix} \frac{1}{2}x& +&y&=&-2\\ x& -&2y & =&8 \end{matrix}\right.$ Determinant $D$ consists of the $x$ and $y$ coefficients. $D=\begin{vmatrix} \frac{1}{2}&1 \\ 1& -2 \end{vmatrix}=(\frac{1}{2})(-2)-(1)(1)=-1-1=-2$ For determinant $D_x$ replace the $x−$ coefficients with the constants. $D_x=\begin{vmatrix} -2&1 \\ 8& -2 \end{vmatrix}=(-2)(-2)-(8)(1)=4-8=-4$ For determinant $D_y$ replace the $y−$ coefficients with the constants. $D_y=\begin{vmatrix} \frac{1}{2}&-2 \\ 1& 8 \end{vmatrix}=(\frac{1}{2})(8)-(1)(-2)=4+2=6$ By using Cramer's rule we have. $x=\dfrac{D_x}{D}=\dfrac{-4}{-2}=2$ and $y=\dfrac{D_y}{D}=\dfrac{6}{-2}=-3$ Hence, the solution set is $(x,y) =(2,-3)$.
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.