Solve the system of equations by elimination. 2x + 3y - 2z = 10 3x - 2y - z = -6 -3x - 2y - 2z = 10 ____

Eliminate x: First, let's add the second and third equations to eliminate x. (3x - 2y - z) + (-3x - 2y - 2z) = -6 + 10 3x - 2y - z - 3x - 2y - 2z = 4 -4y - 3z = 4Eliminate z: Now, let's multiply the first equation by 3 and the second equation by 2 so we can eliminate z by adding them. (2x + 3y - 2z) * 3 = 10 * 3 (3x - 2y - z) * 2 = -6 * 2 6x + 9y - 6z = 30 6x - 4y - 2z = -12Eliminate z: Add the modified equations to eliminate z. (6x + 9y - 6z) + (6x - 4y - 2z) = 30 + (-12) 6x + 9y - 6z + 6x - 4y - 2z = 18 12x + 5y - 8z = 18Eliminate z: Now we have two new equations without z: -4y - 3z = 4 12x + 5y - 8z = 18 Let's multiply the first equation by 2 and add it to the second equation to eliminate z. (-4y - 3z) * 2 = 4 * 2 -8y - 6z = 8 12x + 5y - 8z = 18Add the modified equations to eliminate z. (-8y - 6z) + (12x + 5y - 8z) = 8 + 18 -8y - 6z + 12x + 5y - 8z = 26 12x - 3y - 14z = 26

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Solve the system of equations by elimination. $\newline$ $2x + 3y - 2z = 10$ $\newline$ $3x - 2y - z = -6$ $\newline$ $-3x - 2y - 2z = 10$

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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 + 3y - 2z = 10

\newline

3x - 2y - z = -6

\newline

-3x - 2y - 2z = 10

Eliminate x: First, let's add the second and third equations to eliminate x. $\newline$ $(3x - 2y - z) + (-3x - 2y - 2z) = -6 + 10$ $\newline$ $3x - 2y - z - 3x - 2y - 2z = 4$ $\newline$ $-4y - 3z = 4$
Eliminate z: Now, let's multiply the first equation by $3$ and the second equation by $2$ so we can eliminate $z$ by adding them. $\newline$ $(2x + 3y - 2z) \times 3 = 10 \times 3$ $\newline$ $(3x - 2y - z) \times 2 = -6 \times 2$ $\newline$ $6x + 9y - 6z = 30$ $\newline$ $6x - 4y - 2z = -12$
Eliminate z: Add the modified equations to eliminate z. $\newline$ $(6x + 9y - 6z) + (6x - 4y - 2z) = 30 + (-12)$ $\newline$ $6x + 9y - 6z + 6x - 4y - 2z = 18$ $\newline$ $12x + 5y - 8z = 18$
Eliminate z: Now we have two new equations without z: $\newline$ $-4y - 3z = 4$ $\newline$ $12x + 5y - 8z = 18$ $\newline$ Let's multiply the first equation by $2$ and add it to the second equation to eliminate z. $\newline$ $(-4y - 3z) \times 2 = 4 \times 2$ $\newline$ $-8y - 6z = 8$ $\newline$ $12x + 5y - 8z = 18$
Eliminate z: Now we have two new equations without z: $\newline$ $-4y - 3z = 4$ $\newline$ $12x + 5y - 8z = 18$ $\newline$ Let's multiply the first equation by $2$ and add it to the second equation to eliminate z. $\newline$ $(-4y - 3z) \times 2 = 4 \times 2$ $\newline$ $-8y - 6z = 8$ $\newline$ $12x + 5y - 8z = 18$ Add the modified equations to eliminate z. $\newline$ $(-8y - 6z) + (12x + 5y - 8z) = 8 + 18$ $\newline$ $-8y - 6z + 12x + 5y - 8z = 26$ $\newline$ $12x - 3y - 14z = 26$