Solve the system of equations by substitution. 3x - 2y + 2z = -14 x - y - 3z = -16 x = -6 ____

Substitute x = -6: Substitute x = -6 into the second equation, x - y - 3z = -16. -6 - y - 3z = -16. Now, solve for y. -y - 3z = -16 + 6. -y - 3z = -10. y + 3z = 10. Substitute x = -6: Substitute x = -6 into the first equation, 3x - 2y + 2z = -14. 3(-6) - 2y + 2z = -14. -18 - 2y + 2z = -14. Now, solve for y. -2y + 2z = -14 + 18. -2y + 2z = 4. Divide by -2. y - z = -2. y = z - 2. Solve for z: We have two expressions for y: y + 3z = 10 and y = z - 2. Set them equal to each other to solve for z. z - 2 + 3z = 10. Combine like terms. 4z - 2 = 10. Add 2 to both sides. 4z = 12. Divide by 4. z = 3. Find y: Substitute z = 3 into y = z - 2 to find y. y = 3 - 2. y = 1. Final Solution: We already have x = -6 and z = 3, and we just found y = 1. The solution is (-6, 1, 3).

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Solve the system of equations by substitution. $\newline$ $3x - 2y + 2z = -14$ $\newline$ $x - y - 3z = -16$ $\newline$ $x = -6$

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Algebra 2

Solve a system of equations in three variables using substitution

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Q. Solve the system of equations by substitution.

\newline

3x - 2y + 2z = -14

\newline

x - y - 3z = -16

\newline

x = -6

Substitute $x = -6$ : Substitute $x = -6$ into the second equation, $x - y - 3z = -16$ . $\newline$ $-6 - y - 3z = -16$ . $\newline$ Now, solve for y. $\newline$ $-y - 3z = -16 + 6$ . $\newline$ $-y - 3z = -10$ . $\newline$ $y + 3z = 10$ .
Substitute $x = -6$ : Substitute $x = -6$ into the first equation, $3x - 2y + 2z = -14$ .
$3(-6) - 2y + 2z = -14$ .
$-18 - 2y + 2z = -14$ .
Now, solve for $y$ .
$-2y + 2z = -14 + 18$ .
$-2y + 2z = 4$ .
Divide by $-2$ .
$y - z = -2$ .
$x = -6$ $0$ .
Solve for z: We have two expressions for y: $y + 3z = 10$ and $y = z - 2$ . Set them equal to each other to solve for z. $z - 2 + 3z = 10$ . Combine like terms. $4z - 2 = 10$ . Add $2$ to both sides. $4z = 12$ . Divide by $4$ . $z = 3$ .
Find $y$ : Substitute $z = 3$ into $y = z - 2$ to find $y$ . $\newline$ $y = 3 - 2$ . $\newline$ $y = 1$ .
Final Solution: We already have $x = -6$ and $z = 3$ , and we just found $y = 1$ . The solution is $(-6, 1, 3)$ .