Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 3 - Linear Systems - Chapter Review - Page 186: 25

Answer

Our solution is $x = 6$, $y = 0$, and $z = -2$ or $(6, 0, -2)$

Work Step by Step

Use the expression given for $x$ to substitute into the first two equations: $3(-3z) + y - 2z = 22$ $-3z + 5y + z = 4$ Group like terms: $y + (-9z - 2z) = 22$ $5y + (-3z + z) = 4$ Combine like terms: $y + (-11z) = 22$ $5y + (-2z) = 4$ Simplify: $y - 11z = 22$ $5y - 2z = 4$ We want to modify these equations such that one variable in both equations is the same but differing in sign. We can accomplish this by multiplying the first equation by $-5$: $-5y + 55z = -110$ $5y - 2z = 4$ Combine the two equations by adding them together: $53z = -106$ Divide both sides of the equation by $53$: $z = -2$ Plug this value for $z$ into one of the modified equations to find $y$: $5y - 2(-2) = 4$ Multiply to simplify: $5y + 4 = 4$ Subtract $4$ on both sides of the equation: $5y = 0$ Divide both sides by $5$: $y = 0$ Now that we have values for both $y$ and $z$, we can plug them into one of the original equations to find the value of $x$. Let's use the second equation: $x + 5(0) + (-2) = 4$ Multiply to simplify: $x + 0 - 2 = 4$ Combine like terms: $x - 2 = 4$ Add $2$ to each side of the equation: $x = 6$ Our solution is $x = 6$, $y = 0$, and $z = -2$.
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