College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 5 - Systems of Equations and Inequalities - Exercise Set 5.2 - Page 537: 12


$\{ (1,1,2) \}$

Work Step by Step

Plan (see p.533): 1. Eliminate one variable and arrive at a system of two equations in two variables. 2. Solve the system of two equations in two variables. 3. Back-substitute the solutions of (2) to find the eliminated varable in (1) ------------ (1.) Perform elimination by addition method (eliminate $x$) by $Eq.1-Eq.1 \Rightarrow$ $(1-1)x+(2-0)y+(-1-1)z=1-3$ $Eq.3-2\times Eq.1 \Rightarrow$ $(2-2)x+(-1-0)y+(1-2)z=3-6$ $\left[\begin{array}{ll} 2y-2z=-2 & /\div 2\\ -y-z=-3 & /\times(-1) \end{array}\right.$ , $\left[\begin{array}{ll} y-z=-1 & I\\ y+z=3 & II \end{array}\right.$ ..., (2.) ... $I+II$ eliminates $z$: $2y=2$ $y=1$ Back-substitute into $y+z=3$ $1+z=3$ $z=2$ (3.) Back-substitute into one of the initial equations: $x+z=3$ $x+2=3$ $x=1$ Solution set: : $\{ (1,1,2) \}$
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