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Solve the system of equations by elimination. $\newline$ $-3x + y + z = 5$ $\newline$ $-2x - y + 2z = -19$ $\newline$ $2x + 2y - z = 18$

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Solve a system of equations in three variables using elimination

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Q. Solve the system of equations by elimination.

\newline

-3x + y + z = 5

\newline

-2x - y + 2z = -19

\newline

2x + 2y - z = 18

Combine equations to eliminate y: Combine the first and second equations to eliminate y. $\newline$ $-3x + y + z = 5$ $\newline$ $-2x - y + 2z = -19$ $\newline$ Add them together: $\newline$ $-3x + y + z + (-2x - y + 2z) = 5 - 19$ $\newline$ $-5x + 3z = -14$
Combine equations to eliminate $y$ : Combine the second and third equations to eliminate $y$ .
$-2x - y + 2z = -19$
$2x + 2y - z = 18$
Add them together:
$-2x - y + 2z + (2x + 2y - z) = -19 + 18$
$z = -1$
Substitute $z$ to find $x$ : Substitute $z = -1$ into $-5x + 3z = -14$ to find $x$ .
$-5x + 3(-1) = -14$
$-5x - 3 = -14$
$-5x = -14 + 3$
$-5x = -11$
$x = -11 / -5$
$x$ $0$
Substitute $z$ to find $y$ : Substitute $z = -1$ into the first equation to find $y$ .
$-3x + y + z = 5$
$-3(\frac{11}{5}) + y + (-1) = 5$
$-\frac{33}{5} + y - 1 = 5$
$y = 5 + \frac{33}{5} + 1$
$y = 5 + \frac{33}{5} + \frac{5}{5}$
$y = (\frac{25 + 33 + 5}{5})$
$y$ $0$

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