Solve the system of equations by elimination. 2x + y + z = -4 -x + y - z = 4 2x + 3y + z = -2 ____

Combine Equations to Eliminate z: Combine the first and second equations to eliminate z. (2x + y + z) + (-x + y - z) = -4 + 4 2x + y + z - x + y - z = 0 x + 2y = 0Combine Equations to Simplify: Combine the first and third equations to eliminate z. (2x + y + z) + (2x + 3y + z) = -4 + (-2) 2x + y + z + 2x + 3y + z = -6 4x + 4y = -6Divide and Simplify: Divide the equation from the previous step by 4 to simplify. 4x + 4y = -6 x + y = -6/4 x + y = -3/2Substitute x into Equation: Substitute x = -2y from the equation x + 2y = 0 into x + y = -3/2. (-2y) + y = -3/2 -y = -3/2 y = 3/2Find x: Substitute y = 3/2 into x + 2y = 0 to find x. x + 2(3/2) = 0 x + 3 = 0 x = -3Find z: Substitute x = -3 and y = 3/2 into the first original equation to find z. 2(-3) + (3/2) + z = -4 -6 + 3/2 + z = -4 -12/2 + 3/2 + z = -4 -9/2 + z = -4 z = -4 + 9/2 z = -8/2 + 9/2 z = 1/2

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Solve the system of equations by elimination. $\newline$ $2x + y + z = -4$ $\newline$ $-x + y - z = 4$ $\newline$ $2x + 3y + z = -2$

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Algebra 2

Solve a system of equations in three variables using elimination

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Q. Solve the system of equations by elimination.

\newline

2x + y + z = -4

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-x + y - z = 4

\newline

2x + 3y + z = -2

Combine Equations to Eliminate $z$ : Combine the first and second equations to eliminate $z$ . $\$2x + y + z$ + (-x + y - z) = $-4$ + $4$ \) $2x + y + z - x + y - z = 0$ $x + 2y = 0$
Combine Equations to Simplify: Combine the first and third equations to eliminate $z$ . $\$2x + y + z$ + $2x + 3y + z$ = $-4$ + ( $-2$ )\) $2x + y + z + 2x + 3y + z = -6$ $4x + 4y = -6$
Divide and Simplify: Divide the equation from the previous step by $4$ to simplify. $\newline$ $4x + 4y = -6$ $\newline$ $x + y = -\frac{6}{4}$ $\newline$ $x + y = -\frac{3}{2}$
Substitute $x$ into Equation: Substitute $x = -2y$ from the equation $x + 2y = 0$ into $x + y = -\frac{3}{2}$ . $\newline$ $(-2y) + y = -\frac{3}{2}$ $\newline$ $-y = -\frac{3}{2}$ $\newline$ $y = \frac{3}{2}$
Find $x$ : Substitute $y = \frac{3}{2}$ into $x + 2y = 0$ to find $x$ .
$x + 2\left(\frac{3}{2}\right) = 0$
$x + 3 = 0$
$x = -3$
Find $z$ : Substitute $x = -3$ and $y = \frac{3}{2}$ into the first original equation to find $z$ .
$2(-3) + \left(\frac{3}{2}\right) + z = -4$
$-6 + \frac{3}{2} + z = -4$
$-\frac{12}{2} + \frac{3}{2} + z = -4$
$-\frac{9}{2} + z = -4$
$z = -4 + \frac{9}{2}$
$z = -\frac{8}{2} + \frac{9}{2}$
$x = -3$ $0$