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: 14


$\{ (0,0,4) \}$

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) by $Eq.$2 already is without x. $-3Eq.1+Eq.3 \Rightarrow$ $(-3+3)x+(-2-9)y+(9-15)z=-60+36$ $\left[\begin{array}{ll} y-4z=-16 & I\\ -11y-6z=-24 & II \end{array}\right.$ , ({\it 2}.) ... $11\cdot I+II$ eliminates $y$: $(-44-6)z=-176-24$ $-50z=-200$ $z=4$ Back-substitute into $y-4z=-16$ $y-4(4)=-16$ $y=0$ (3.) Back-substitute into one of the initial equations: $x+3y+5z=20$ $1+0+20=20$ $x=0$ Solution set: : $\{ (0,0,4) \}$
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