Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 946: 46


The value of $x$ is $=26$

Work Step by Step

Consider, $\left| \begin{matrix} x+3 & -6 \\ x-2 & -4 \\ \end{matrix} \right|=28$ Calculate the above determinant as follows: $\left| \begin{matrix} x+3 & -6 \\ x-2 & -4 \\ \end{matrix} \right|=28$ This implies that, $-4\left( x+3 \right)+6\left( x-2 \right)=28$ Simplify the above equation, $-4x-12+6x-12=28$ Or, $2x=52$ Divide by $2$. Now, $x=26$. Hence, $x=26$
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