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: 11


$\{ (3,1,5) \}$

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 $Eq.1-2\times Eq.2 \Rightarrow$ $(2-2)x+(-4-4)y+(3+2)z=17-0$ $Eq.3-4\times Eq.2 \Rightarrow$ $(4-4)x+(-1-8)y+(-1+4)z=6-0$ $\left[\begin{array}{ll} -8y+5z=17 & \\ -9y+3z=6 & \div(-3) \end{array}\right.$ , $\left[\begin{array}{ll} -8y+5z=17 & I\\ 3y-z=-2 & II \end{array}\right.$ ..., (2.) ... $I+5\times II$ eliminates $z$: $(-8+15)z=17-10$ $7y=7$ $y=1$ Back-substitute into $3y-z=-2$ $3(3)-z=-2$ $3+2=z$ $z=5$ (3.) Back-substitute into one of the initial equations: $x+2y-z=0$ $x+2(1)-(5)=0$ $x=3$ Solution set: : $\{ (3,1,5) \}$
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