Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 3 - Section 3.3 - Systems of Linear Equations in Three Variables - Exercise Set - Page 216: 21

Answer

$\{\left(\frac{1}{2},\frac{1}{3},-1 \right ) \}$.

Work Step by Step

The given equations are $\Rightarrow 3(2x+y)+5z=-1$ Apply distributive property $\Rightarrow 6x+3y+5z=-1$ ......(1) $\Rightarrow 2(x-3y+4z)=-9$ Apply distributive property $\Rightarrow 2x-6y+8z=-9$......(2) $\Rightarrow 4(1+x)=-3(z-3y)$ Apply distributive property $\Rightarrow 4+4x=-3z+9y$ Add $+3z-9y-4$ to both sides. $\Rightarrow 4+4x+3z-9y-4=-3z+9y+3z-9y-4$ Add like terms. $\Rightarrow 4x-9y+3z=-4$......(3) Multiply equation (1) by $2$. $\Rightarrow 12x+6y+10z=-2$ ......(4) Now add equations (2) and (4). $\Rightarrow 2x-6y+8z+12x+6y+10z=-9-2$ Add like terms. $\Rightarrow 14x+18z=-11$ ...... (5) Multiply equation (1) by $3$. $\Rightarrow 18x+9y+15z=-3$.......(6) Now add equation (3) and (6). $\Rightarrow 4x-9y+3z+18x+9y+15z=-4-3$ Add like terms. $\Rightarrow 22x+18z=-7$......(7) Subtract equation (5) from equation (7). $\Rightarrow 22x+18z-(14x+18z)=-7-(-11)$ Simplify. $\Rightarrow 22x+18z-14x-18z=-7+11$ $\Rightarrow 8x=4$ Divide both sides by $8$. $\Rightarrow x=\frac{1}{2}$ Substitute the value of $x$ into equation (5). $\Rightarrow 14(\frac{1}{2})+18z=-11$ Simplify. $\Rightarrow 7+18z=-11$ Subtract $7$ from both sides. $\Rightarrow 7+18z-7=-11-7$ Simplify. $\Rightarrow 18z=-18$ Divide both sides by $18$. $\Rightarrow z=-1$ Substitute the values of $x$ and $z$ into equation (1). $\Rightarrow 6(\frac{1}{2})+3y+5(-1)=-1$ Simplify. $\Rightarrow 3+3y-5=-1$ $\Rightarrow 3y-2=-1$ Add $2$ to both sides. $\Rightarrow 3y-2+2=-1+2$ Simplify. $\Rightarrow 3y=1$ Divide both sides by $3$. $\Rightarrow y=\frac{1}{3}$ The solution set is $\{(x,y,z)\}=\{\left(\frac{1}{2},\frac{1}{3},-1 \right ) \}$.
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