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: 22

Answer

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

Work Step by Step

The given equations are $\Rightarrow 7z-3=2(x-3y)$ Apply distributive property $\Rightarrow 7z-3=2x-6y$ Add $-2x+6y+3$ to both sides. $\Rightarrow 7z-3-2x+6y+3=2x-6y-2x+6y+3$ Add like terms. $\Rightarrow -2x+6y+7z=3$......(1) $\Rightarrow 5y+3z-7=4x$ Add $-4x+7$ to both sides. $\Rightarrow 5y+3z-7-4x+7=4x-4x+7$ Add like terms. $\Rightarrow -4x+5y+3z=7$......(2) $\Rightarrow 4+5z=3(2x-y)$ Apply distributive property $\Rightarrow 4+5z=6x-3y$ Add $-6x+3y-4$ to both sides. $\Rightarrow 4+5z-6x+3y-4=6x-3y-6x+3y-4$ Add like terms. $\Rightarrow -6x+3y+5z=-4$.....(3) Multiply equation (1) by $-2$. $\Rightarrow 4x-12y-14z=-6$......(4) Now add equations (2) and (4). $\Rightarrow -4x+5y+3z+4x-12y-14z=7-6$ Add like terms. $\Rightarrow -7y-11z=1$...... (5) Multiply equation (1) by $-3$. $\Rightarrow 6x-18y-21z=-9$......(6) Now add equations (3) and (6). $\Rightarrow -6x+3y+5z+6x-18y-21z=-4-9$ Add like terms. $\Rightarrow -15y-16z=-13$......(7) Multiply equation (5) by $-15$ and equation (7) by $7$ and then add. $\Rightarrow -15(-7y-11z)+7(-15y-16z)=-15(1)+7(-13)$ Apply distributive property. $\Rightarrow 105y+165z-105y-112z=-15-91$ Add like terms. $\Rightarrow 53z=-106$ Divide both sides by $53$. $\Rightarrow \frac{53z}{53}=\frac{-106}{53}$ Simplify. $\Rightarrow z=-2$ Substitute the value of $z$ into equation (5). $\Rightarrow -7y-11(-2)=1$ Simplify. $\Rightarrow -7y+22=1$ Subtract $22$ from both sides. $\Rightarrow -7y+22-22=1-22$ Simplify. $\Rightarrow -7y=-21$ Divide both sides by $-7$. $\Rightarrow y=3$ Substitute the values of $y$ and $z$ into equation (1). $\Rightarrow -2x+6(3)+7(-2)=3$ Simplify. $\Rightarrow -2x+18-14=3$ $\Rightarrow -2x+4=3$ Subtract $4$ from both sides. $\Rightarrow -2x+4-4=3-4$ $\Rightarrow -2x=-1$ Divide both sides by $-2$. $\Rightarrow x=\frac{1}{2}$ The solution set is $\{(x,y,z)\}=\{\left(\frac{1}{2},3,-2 \right ) \}$.
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