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: 24

Answer

$(5, 2, 2)$

Work Step by Step

Label the original equations first: 1. $13 = 3x - y$ 2. $4y - 3x + 2z = -3$ 3. $z = 2x - 4y$ Substitute the expression for $z$ given by equation $3$ into equation $2$: $4y - 3x + 2(2x - 4y) = -3$ Use distributive property to simplify: $4y - 3x + 4x - 8y = -3$ Combine like terms: $x - 4y = -3$ Transform equation $4$ so that we can add it to equation $1$. To do this, multiply equation $4$ by $-3$: $-3(x - 4y) = -3(-3)$ Distribute and multiply to simplify: $-3x + 12y = 9$ Set up a system of equations made up of equations $1$ and $5$: $3x - y = 13$ $-3x + 12y = 9$ Add the two equations together: $11y = 22$ Divide each side of the equation by $11$: $y = 2$ Substitute this value for $y$ into equation $1$ to find the value of $x$: $13 = 3x - 2$ Add $2$ to each side of the equation to move constants to one side of the equation: $3x = 15$ Divide each side by $3$ to solve for $x$: $x = 5$ Substitute the values for $x$ and $y$ into equation $3$ to find the value of $z$: $z = 2(5) - 4(2)$ Multiply to simplify: $z = 10 - 8$ Subtract to solve for $z$: $z = 2$ The solution is $(5, 2, 2)$. To check the solution, plug in the three values into one of the original equations. Use equation $2$: $4(2) - 3(5) + 2(2) = -3$ $8 - 15 + 4 = -3$ $-3 = -3$ The sides are equal to one another; therefore, the solution is correct.
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