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: 9


$\{ (1,2,3) \}$

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 $x$) by $2\times Eq.1-3\times Eq.2 \Rightarrow$ $(6-6)x+(4+15)y+(-6-6)z=-4+6$ $4\times Eq.1-3\times Eq3 \Rightarrow$ $(12-12)x+(8+9)y+(-12-12)z=-8-30$ $\left[\begin{array}{ll} 19y-12z=2 & I\\ 17y-24z=-38 & II \end{array}\right.$ ..., (2.) ...$ II-2\times I$ eliminates z: $(17-38)y=-38-4$ $-21y=-42$ $y=2$ Back-substitute into$ 19y-12z=2$ $19(2)-12z=2$ $-12z=2-38$ $z=3$ (3.) Back-substitute into one of the initial equations: $3x+2y-3z=-2$ $3x+2(2)-3(3)=-2$ $3x-5=-2$ $3x=3$ $x=1$ Solution set: : $\{ (1,2,3) \}$
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