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: 17


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

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) ------------ First, write each equation in standard form: Ax+By+Cz=D. $\left[\begin{array}{l} 6x+3y+5z=-1\\ 2x-6y+8z=-9\\ 4+4x=-3z+9y \end{array}\right.$ $\left[\begin{array}{l} 6x+3y+5z=-1\\ 2x-6y+8z=-9\\ 4x-9y+3z==4 \end{array}\right.$ (1.) Perform elimination by addition method (eliminate x) by $\left[\begin{array}{lll} Eq.1.-3\times Eq.2\rightarrow & 21y-19z=26 & I\\ -2\times Eq.2.+Eq.3 & 3y-13z=14 & II \end{array}\right.$ , (2.) ... $(Eq.I - 7\times Eq.II)$ eliminates y, $72z=-72$ $z=-1$ Back-substitute into $21y-19z=26$ $21y-19(-1)=26$ $21y=26-19$ $y=\displaystyle \frac{7}{21}=\frac{1}{3}$ (3.) Back-substitute into one of the initial equations: $6x+3(\displaystyle \frac{1}{3})+5(-1)=-1$ $6x-4=-1$ $6x=3$ $x=\displaystyle \frac{1}{2}$ Solution set: : $\{ (\displaystyle \frac{1}{2},\frac{1}{3},-1) \}$
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