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: 29

Answer

$(1, 3, 2)$

Work Step by Step

Label the original equations: 1. $5x - y + z = 4$ 2. $x + 2y - z = 5$ 3. $2x + 3y - 3z = 5$ Solve equation $1$ for $z$. This will be equation $4$: 4. $z = -5x + y + 4$ Substitute the expression for $z$ given by equation $4$ into equation $2$: $x + 2y - (-5x + y + 4) = 5$ Combine like terms. This will be equation $5$: 5. $6x + y - 4 = 5$ Move constants to the right side of the equation by adding $4$ to both sides of the equation: 5. $6x + y = 9$ Substitute the expression for $x$ given by equation $4$ into equation $3$: $2x + 3y - 3(-5x + y + 4) = 5$ Distribute and multiply to simplify. This will be equation $6$: 6. $2x + 3y + 15x - 3y - 12 = 5$ Combine like terms on the left side of the equation: 6. $17x - 12 = 5$ Move constants to the right side of the equation by subtracting $10$ from both sides of the equation: 6. $17x = 17$ Divide both sides of the equation by $17$ to solve for $x$: $x = 1$ Substitute this value for $x$ into equation $5$: $6(1) + y = 9$ Subtract $6$ from both sides of the equation to solve for $y$: $y = 3$ Substitute the values for $x$ and $y$ into equation $2$ to find the value of $z$: $1 + 2(3) - z = 5$ Multiply to simplify: $1 + 6 - z = 5$ Combine like terms on the left side of the equation: $7 - z = 5$ Subtract $7$ from both sides of the equation to move constants to the right side of the equation: $-z = -2$ Divide both sides of the equation by $-1$ to solve for $z$: $z = 2$ The solution is $(1, 3, 2)$. To check the solution, plug in the three values into one of the original equations. Use equation $3$: $2(1) + 3(3) - 3(2) = 5$ $2 + 9 - 6 = 5$ $5 = 5$ The sides are equal to one another; therefore, the solution is correct.
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