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: 14

Answer

$(0, 3, -2)$

Work Step by Step

Label the original equations first: 1. $x - y + 2z = -7$ 2. $y + z = 1$ 3. $x = 2y + 3z$ Rewrite equation $3$ so that all variables are on the left side of the equation: 4. $x - 2y - 3z = 0$ Modify one of the equations so that the $x$ term can be eliminated. This will allow us to be able to add this equation to equation $2$ to eliminate one variable and be able to solve for the other. Multiply equation $1$ by $-1$: 5. $-1(x - y + 2z) = -1(-7)$ Use distributive property: 5. $-x + y - 2z = 7$ Combine equations $4$ and $5$: 4. $x - 2y - 3z = 0$ 5. $-x + y - 2z = 7$ Add the equations together: 7. $-y - 5z = 7$ Combine equations $2$ and $7$: 2. $y + z = 1$ 7. $-y - 5z = 7$ Add the equations: $-4z = 8$ Divide both sides of the equation by $-4$: $z = -2$ Substitute this value into equation $2$ to solve for $y$: 2. $y + (-2) = 1$ Add $2$ to each side of the equation: $y = 3$ Substitute the values of $y$ and $z$ into one of the original equations. Let's use equation $3$: 3. $x = 2(3) + 3(-2)$ Multiply to simplify: $x = 6 - 6$ Combine like terms to solve for $x$: $x = 0$ The solution is $(0, 3, -2)$. To check the solution, plug in the three values into one of the original equations. Use equation $1$: $0 - 3 + 2(-2) = -7$ $0 - 3 - 4 = -7$ $-7 = -7$ The sides are equal to one another; therefore, the solution is correct.
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