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

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

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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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Precalculus

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. $\newline$ $-3(0) + 2(7) - (-6) = 20$ $\newline$ $0 + 14 + 6 = 20$ $\newline$ $20 = 20$