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 171: 10

Answer

$(0, 2, -3)$

Work Step by Step

Label the original equations first: 1. $x - y - 2z = 4$ 2. $-x + 2y + z = 1$ 3. $-x + y - 3z = 11$ The first step is to combine equations $1$ and $3$ to see if one variable can be eliminated: 1. $x - y - 2z = 4$ 3. $-x + y - 3z = 11$ Add the equations together: $-5z = 15$ Divide both sides of the equation by $-5$ to solve for $z$: $z = -3$ Substitute this value for $z$ into both equations $1$ and $2$ to eliminate the $x$ variable: 1. $x - y - 2(-3) = 4$ 2. $-x + 2y + (-3) = 1$ Multiply to simplify: $x - y + 6 = 4$ $-x + 2y - 3 = 1$ Move constants to the right side of the equation: $x - y = -2$ $-x + 2y = 4$ Add the equations together: $y = 2$ Substitute these values for $y$ and $z$ into one of the original equations to solve for $x$. Use equation $1$: 1. $x - 2 - 2(-3) = 4$ Multiply to simplify: $x - 2 + 6 = 4$ Combine like terms: $x + 4 = 4$ Subtract $4$ from each side of the equation: $x = 0$ The solution is $(0, 2, -3)$.
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