Precalculus (6th Edition) Blitzer

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

Chapter 7 - Section 7.6 - Linear Programming - Exercise Set - Page 874: 42

Answer

The solution set is $\left\{ \left( -2,1,4,3 \right) \right\}$

Work Step by Step

We have the system of equations, $\begin{align} & w-x+2y-2z=-1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & x-\frac{1}{3}y+z=\frac{8}{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \\ & y-z=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{III} \right) \\ & z=3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{IV} \right) \end{align}$ Substitute the value of z from the 4th equation to the 3rd equation. So, $\begin{align} & y-z=1 \\ & y-3=1 \\ & y=4 \end{align}$ Now, put the value of $y=4$ and $z=3$ into equation (II) to find the value of x: $\begin{align} & x-\frac{1}{3}y+z=\frac{8}{3} \\ & x-\frac{1}{3}\left( 4 \right)+3=\frac{8}{3} \\ & x-\frac{4}{3}+3=\frac{8}{3} \\ & x-\frac{4}{3}+\frac{9}{3}=\frac{8}{3} \end{align}$ Finally, $\begin{align} & x=\frac{8}{3}-\frac{5}{3} \\ & x=1 \\ \end{align}$ Now, put the value of $x=1$ , $y=4$ and $z=3$ into equation (I) to find the value of w: $\begin{align} & w-x+2y-2z=-1 \\ & w-1+2\left( 4 \right)-2\left( 3 \right)=-1 \\ & w-1+8-6=-1 \\ & w+1=-1 \end{align}$ Finally, $\begin{align} & w=-1-1 \\ & w=-2 \\ \end{align}$ Hence, the solution set is $\left\{ \left( -2,1,4,3 \right) \right\}$.
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