Precalculus (6th Edition) Blitzer

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

Chapter 7 - Section 7.2 - Systems of Linear Equations in Three Variables - Concept and Vocabulary Check - Page 829: 1


A solution of a system of linear equations in three variables is an ordered triple of real numbers that satisfies all equations of the system.

Work Step by Step

To satisfy the system of linear equations, an ordered triple must satisfy all equations of the system. Example: Let us take an example of a system of equations as, $\left\{ \begin{align} & x+2y+3z=-2 \\ & 3x+3y+10z=-2 \\ & 2y-5z=6 \end{align} \right.$ Equation $2y-5z=6$ has only two variables. Now, consider equation $x+2y+3z=-2$ and $3x+3y+10z=-2$, and eliminate z, to get $2y-5z=6$ and then eliminate another variable. Then, multiply equation $x+2y+3z=-2$ by $3$ and equation $3x+3y+10z=-2$ by $1$ and get; $\begin{align} & \left( x+2y+3z \right)\times 3=-2\times 3 \\ & 3x+6y+9z=-6 \end{align}$ And, $3x+3y+10z=-2$. Now, subtract equation $3x+6y+9z=-6$ and $3x+3y+10z=-2$, to get; $\begin{align} & \left( 3x+6y+9z \right)-\left( 3x+3y+10z \right)=\left( -6 \right)-\left( -2 \right) \\ & 3y-z=-4 \end{align}$ Now, solve equations $2y-5z=6$ and $3y-z=-4$ and obtain the values of z and y. The values so obtained: $\begin{align} & z=-2 \\ & y=-2 \end{align}$ Put the values of z and y in equation $x+2y+3z=-2$ and get the value of $x$ as, $\begin{align} & x+2y+3z=-2 \\ & x+2\left( -2 \right)+3\left( -2 \right)=-2 \\ & x-4-6=-2 \\ & x=8 \end{align}$ Values of x, y, and z are $\left( 8,-2,-2 \right)$. Get the ordered triple solution set $\left( x,y,z \right)$ is $\left( 8,-2,-2 \right)$. Since, the solution set $\left( 8,-2,-2 \right)$ is calculated by the use of the given system of equation, so it must satisfy the system of equations.
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