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 - Lesson Check - Page 171: 2

Answer

$(6, 0, -2)$

Work Step by Step

Label the original equations first: 1. $3x + y - 2z = 22$ 2. $x + 5y +z = 4$ 3. $x = -3z$ The first step is to substitute the expression for $x$ into both the first and second equations to eliminate one variable. Label these the fourth and fifth equations: 4. $3(-3z) + y - 2z = 22$ 5. $-3z + 5y +z = 4$ Multiply to simplify. 4. $-9z + y - 2z = 22$ 5. $-3z + 5y +z = 4$ Combine like terms: 4. $-11z + y = 22$ 5. $-2z + 5y = 4$ Modify these two equations such that one variable is the same in both equations but differs only in sign. Multiply the fourth equation by $-5$: 6. $55z - 5y = -110$ Combine this equation and the fifth equation to eliminate the $x$ variable: 5. $-2z + 5y = 4$ 6. $55z - 5y = -110$ Add the equations together: $53z = -106$ Divide both sides of the equation by $53$: $z = -2$ Substitute this $z$ value into the third equation to find the value of $x$: 3. $x = -3(-2)$ Multiply to solve for $x$: $x = 6$ Substitute the $x$ and $z$ values into the second equation to find $y$: 2. $6 + 5y +(-2) = 4$ Combine like terms on the left side of the equation: $$5y + 4 = 4$$ Subtract $4$ from each side of the equation: $5y = 0$ Divide each side of the equation by $5$: $$y = 0$$ Our solution is $(6, 0, -2)$.
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