Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 6 - Section 6.5 - Synthetic Division and the Remainder Theorem - Exercise Set - Page 455: 56

Answer

$\{(5,3)\}$.

Work Step by Step

The given system of equations is $\left\{\begin{matrix} 7x& -6y&=&17\\ 3x& +y & =&18 \end{matrix}\right.$ Determinant $D$ consists of the $x$ and $y$ coefficients. $D=\begin{vmatrix} 7& -6 \\ 3& 1 \end{vmatrix}=(7)(1)-(3)(-6)=7+18=25$ For determinant $D_x$ replace the $x−$ coefficients with the constants. $D_x=\begin{vmatrix} 17& -6 \\ 18& 1 \end{vmatrix}=(17)(1)-(18)(-6)=17+108=125$ For determinant $D_y$ replace the $y−$ coefficients with the constants. $D_y=\begin{vmatrix} 7& 17 \\ 3& 18 \end{vmatrix}=(7)(18)-(3)(17)=126-51=75$ By using Cramer's rule we have. $x=\frac{D_x}{D}=\frac{125}{25}=5$ and $y=\frac{D_y}{D}=\frac{75}{25}=3$ Hence, the solution set is $\{(x,y)\} =\{(5,3)\}$.
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