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Solve the system of equations by substitution. $\newline$ $z = 4$ $\newline$ $3x + y - 3z = -1$ $\newline$ $x + y - z = -1$

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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

z = 4

\newline

3x + y - 3z = -1

\newline

x + y - z = -1

Substitute $z = 4$ : Substitute $z = 4$ into $3x + y - 3z = -1$ . $\newline$ $3x + y - 3\times4 = -1$ $\newline$ $3x + y - 12 = -1$ $\newline$ $3x + y = 11$
Substitute $z = 4$ : Substitute $z = 4$ into $x + y - z = -1$ . $\newline$ $x + y - 4 = -1$ $\newline$ $x + y = 3$
Two Equations: Now we have two equations: $\newline$ $3x + y = 11$ $\newline$ $x + y = 3$ $\newline$ Subtract the second equation from the first to eliminate $y$ . $\newline$ $(3x + y) - (x + y) = 11 - 3$ $\newline$ $3x + y - x - y = 8$ $\newline$ $2x = 8$
Eliminate $y$ : Solve for $x$ . $2x = 8$ $x = \frac{8}{2}$ $x = 4$
Solve for $x$ : Substitute $x = 4$ into $x + y = 3$ . $4 + y = 3$ $y = 3 - 4$ $y = -1$
Substitute $x = 4$ : We have found the values: $\newline$ $x = 4$ $\newline$ $y = -1$ $\newline$ $z = 4$

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