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: 13


$\{ (1,0,-3) \}$

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 z): $Eq.1$ already is without z. $Eq.2+Eq.3 \Rightarrow$ $(3+1)x+(2+1)y+(1-1)z=4+0$ $\left[\begin{array}{ll} 2x+y=2 & I\\ 4x+3y=4 & II \end{array}\right.$ , (2.) ... $-3I+II$ eliminates $y$: $(-6+4)x=-6+4$ $x=1$ Back-substitute into $2x+y=2$ $2(1)+y=2$ $y=0$ (3.) Back-substitute into one of the initial equations: $x+y-z=4$ $1+0-z=4$ $z=-3$ Solution set: : $\{ (1,0,-3) \}$
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