Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 3 - Linear Systems - 3-5 Systems With Three Variables - Practice and Problem-Solving Exercises - Page 172: 40

Answer

The solution is $(-1, 2, 0)$.

Work Step by Step

Label the original equations: 1. $4y + 2x = 6 - 3z$ 2. $x + z - 2y = -5$ 3. $x - 2z = 3y - 7$ The first step is to rewrite these equations in a more standard form: 1. $2x + 4y + 3z = 6$ 2. $x - 2y + z = -5$ 3. $x - 3y - 2z = -7$ Now, choose two equations to work with where one variable can be eliminated. Let's choose equations $2$ and $3$. Modify the equations by multiplying them by a non-zero factor so that one variable is the same in both equations but differing in sign so that this variable can be eliminated when the two equations are added together. Multiply equation $2$ by $-1$ and leave equation $3$ as-is: 2. $-1(x - 2y + z) = -1(-5)$ 3. $x - 3y - 2z = -7$ Distribute and multiply to simplify: $-x + 2y - z = 5$ $x - 3y - 2z = -7$ Add the equations. This will become equation $4$: 4. $-y - 3z = -2$ Now, choose another two equations again and modify them. Modify these equations such that the $x$ variable can be eliminated. This modified equation will be equation $5$ and will be added to equation $4$ to try to eliminate another variable. Use equations $1$ and $3$. Multiply equation $3$ by $-2$ and leave equation $1$ as-is: 1. $2x + 4y + 3z = 6$ 3. $-2(x - 3y - 2z) = -2(-7)$ Distribute and multiply to simplify: $2x + 4y + 3z = 6$ $-2x + 6y + 4z = 14$ Add the equations together. This will become equation $5$: 5. $10y + 7z = 20$ Set up a system of equations consisting of equations $4$ and $5$: 4. $-y - 3z = -2$ 5. $10y + 7z = 20$ Modify these equations such that the $y$ variable can be eliminated. Multiply equation $4$ by $10$ and leave equation $5$ as-is: 4. $10(-y - 3z) = 10(-2)$ 5. $10y + 7z = 20$ Distribute and multiply to simplify: $-10y - 30z = -20$ $10y + 7z = 20$ Add the two equations together: $-23z = 0$ Divide both sides by $-23$ to solve for $z$: $z = 0$ Substitute this value for $z$ into equation $4$ to solve for $y$: $-y - 3(0) = -2$ $-y = -2$ $y = 2$ Substitute the values for $y$ and $z$ into one of the original equations to find $x$. Use equation $2$: $x - 2(2) + 0 = -5$ $x - 4 + 0 = -5$ $x - 4 = -5$ Add $4$ to each side of the equation to solve for $x$: $x = -1$ The solution is $(-1, 2, 0)$.
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