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Solve the system of equations by substitution. $\newline$ $-2x - 3y - 3z = 2$ $\newline$ $-x - 2y + 2z = -8$ $\newline$ $y = 4$

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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 = 2

\newline

-x - 2y + 2z = -8

\newline

y = 4

Substitute $y = 4$ : Substitute $y = 4$ into $-2x - 3y - 3z = 2$ .
$-2x - 3(4) - 3z = 2$
$-2x - 12 - 3z = 2$
$-2x - 3z = 14$
Substitute $y = 4$ : Substitute $y = 4$ into $-x - 2y + 2z = -8.$ $\newline$ $-x - 2(4) + 2z = -8$ $\newline$ $-x - 8 + 2z = -8$ $\newline$ $-x + 2z = 0$
Solve for x: Solve $-x + 2z = 0$ for $x$ . $\newline$ $-x + 2z = 0$ $\newline$ $x = 2z$
Substitute $x = 2z$ : Substitute $x = 2z$ into $-2x - 3z = 14$ .
$-2(2z) - 3z = 14$
$-4z - 3z = 14$
$-7z = 14$
Solve for $z$ : Solve $-7z = 14$ for $z$ . $\newline$ $z = \frac{14}{-7}$ $\newline$ $z = -2$
Substitute $z = -2$ : Substitute $z = -2$ into $x = 2z$ . $\newline$ $x = 2(-2)$ $\newline$ $x = -4$

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