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: 10


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

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 $3\times Eq.1-2\times Eq.2 \Rightarrow$ $(6-6)x+(9-4)y+(21+10)z=39+44$ $5\times Eq.1-2\times Eq3 \Rightarrow$ $(10-10)x+(15-14)y+(35+6)z=65+56$ $\left[\begin{array}{ll} 5y+31z=83 & I\\ y+41z=121 & II \end{array}\right.$ ..., (2.) ...$( I-5\times II)$ eliminates $y$: $(31-205)z=83-605$ $-174z=-522$ $z=3$ Back-substitute into $\ \ y+41z=121$ $y+41(3)=121$ $y=121-123$ $y=-2$ (3.) Back-substitute into one of the initial equations: $2x+3y+7z=13$ $2x+3(-2)+7(3)=13$ $2x=13-15$ $x=-1$ Solution set: : $\{ (-1,-2,3) \}$
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