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Solve the system of equations by substitution. $\newline$ $y = 7$ $\newline$ $x + 3y + 3z = 3$ $\newline$ $-3x + 2y - z = 20$

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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

y = 7

\newline

x + 3y + 3z = 3

\newline

-3x + 2y - z = 20

Substitute $y = 7$ : First, let's substitute $y = 7$ into the second equation. $\newline$ $x + 3(7) + 3z = 3$ $\newline$ $x + 21 + 3z = 3$ $\newline$ Now, we'll solve for x. $\newline$ $x + 3z = 3 - 21$ $\newline$ $x + 3z = -18$
Solve for x: Next, we substitute $y = 7$ into the third equation. $-3x + 2(7) - z = 20$ $-3x + 14 - z = 20$ Now, we'll solve for z. $-3x - z = 6$
Substitute $y = 7$ : We have two equations now: $\newline$ $1$ ) $x + 3z = -18$ $\newline$ $2$ ) $-3x - z = 6$ $\newline$ Let's multiply the second equation by $3$ to eliminate $x$ . $\newline$ $3(-3x - z) = 3(6)$ $\newline$ $-9x - 3z = 18$
Solve for z: Now we add the first equation to the new equation we got from the second one. $\newline$ $(x + 3z) + (-9x - 3z) = -18 + 18$ $\newline$ $x - 9x + 3z - 3z = 0$ $\newline$ $-8x = 0$ $\newline$ $x = 0 / -8$ $\newline$ $x = 0$
Eliminate $x$ : Now we'll substitute $x = 0$ back into the first equation to find $z$ . $0 + 3z = -18$ $3z = -18$ $z = -18 / 3$ $z = -6$
Add equations: Finally, we'll substitute $x = 0$ and $z = -6$ back into the second equation to check if it's correct. $\newline$ $-3(0) + 2(7) - (-6) = 20$ $\newline$ $0 + 14 + 6 = 20$ $\newline$ $20 = 20$

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