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Solve the system of equations by elimination. $\newline$ $2x - 3y + 2z = 16$ $\newline$ $2x + 3y + 2z = 16$ $\newline$ $-3x + 2y + 2z = -9$

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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

2x - 3y + 2z = 16

\newline

2x + 3y + 2z = 16

\newline

-3x + 2y + 2z = -9

Eliminate y: Add the first two equations to eliminate y. $\newline$ $(2x - 3y + 2z) + (2x + 3y + 2z) = 16 + 16$ $\newline$ $4x + 4z = 32$ $\newline$ Divide by $4$ to simplify. $\newline$ $x + z = 8$
Find $y=0$ : Subtract the second equation from the first to eliminate $y$ again. $(2x - 3y + 2z) - (2x + 3y + 2z) = 16 - 16$ $-6y = 0$ Divide by $-6$ . $y = 0$
Substitute $y=0$ : Substitute $y = 0$ into the third equation to find $x$ and $z$ .
$-3x + 2(0) + 2z = -9$
$-3x + 2z = -9$
Now we have two equations with $x$ and $z$ :
$1$ ) $x + z = 8$
$2$ ) $-3x + 2z = -9$
Align coefficients of $x$ : Multiply the first equation by $3$ to align with the coefficient of $x$ in the second equation. $\newline$ $3(x + z) = 3(8)$ $\newline$ $3x + 3z = 24$
Eliminate $x$ : Add the new equation to the second equation to eliminate $x$ .
$(3x + 3z) + (-3x + 2z) = 24 + (-9)$
$5z = 15$
Divide by $5$ .
$z = 3$
Find $z=3$ : Substitute $z = 3$ into $x + z = 8$ to find $x$ .
$x + 3 = 8$
$x = 8 - 3$
$x = 5$
Find $x=5$ : We have found the values: $\newline$ $x = 5$ $\newline$ $y = 0$ $\newline$ $z = 3$

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