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Solve the system of equations by substitution. $\newline$ $-2x - 3y + 3z = -14$ $\newline$ $x = -8$ $\newline$ $x + 3y + 3z = 16$

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Solve a system of equations in three variables using substitution

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Q. Solve the system of equations by substitution.

\newline

-2x - 3y + 3z = -14

\newline

x = -8

\newline

x + 3y + 3z = 16

Substitute $x = -8$ : Substitute $x = -8$ into $-2x - 3y + 3z = -14$ .
$-2(-8) - 3y + 3z = -14$
$16 - 3y + 3z = -14$
Solve for $-3y + 3z$ : Subtract $16$ from both sides to solve for $-3y + 3z$ . $\newline$ $-3y + 3z = -14 - 16$ $\newline$ $-3y + 3z = -30$
Substitute $x = -8$ : Substitute $x = -8$ into $x + 3y + 3z = 16$ . $\newline$ $-8 + 3y + 3z = 16$ $\newline$ $3y + 3z = 16 + 8$ $\newline$ $3y + 3z = 24$
Simplify equation: Divide the equation $3y + 3z = 24$ by $3$ to simplify. $\newline$ $y + z = \frac{24}{3}$ $\newline$ $y + z = 8$
Align z terms: Now we have two equations with y and z: $\newline$ $1$ ) $-3y + 3z = -30$ $\newline$ $2$ ) $y + z = 8$ $\newline$ Multiply the second equation by $3$ to align the z terms. $\newline$ $3(y + z) = 3(8)$ $\newline$ $3y + 3z = 24$
Eliminate z: Subtract the new equation from the first one to eliminate z. $\newline$ $(-3y + 3z) - (3y + 3z) = -30 - 24$ $\newline$ $-3y + 3z - 3y - 3z = -54$ $\newline$ $-6y = -54$
Solve for y: Divide both sides by $-6$ to solve for y. $\newline$ $-6y / -6 = -54 / -6$ $\newline$ $y = 9$
Find $z$ : Substitute $y = 9$ into $y + z = 8$ to find $z$ . $\newline$ $9 + z = 8$ $\newline$ $z = 8 - 9$ $\newline$ $z = -1$
Final solution: We have found the values for $x$ , $y$ , and $z$ : $x = -8$ $y = 9$ $z = -1$ The solution is $(-8, 9, -1)$ .

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