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Solve the system of equations by substitution. $\newline$ $-x + y - 2z = -7$ $\newline$ $y = 1$ $\newline$ $x - 2y - z = 3$

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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

-x + y - 2z = -7

\newline

y = 1

\newline

x - 2y - z = 3

Substitute $y = 1$ : Substitute $y = 1$ into $-x + y - 2z = -7$ . $\newline$ $-x + 1 - 2z = -7$ $\newline$ $-x - 2z = -8$
Substitute $y = 1$ : Substitute $y = 1$ into $x - 2y - z = 3$ . $\newline$ $x - 2(1) - z = 3$ $\newline$ $x - 2 - z = 3$ $\newline$ $x - z = 5$
Solve for x: Solve for $x$ from $x - z = 5$ . $\newline$ $x = z + 5$
Substitute $x = z + 5$ : Substitute $x = z + 5$ into $-x - 2z = -8$ .
$-(z + 5) - 2z = -8$
$-z - 5 - 2z = -8$
$-3z - 5 = -8$
Solve for z: Solve for z from $-3z - 5 = -8$ .
$-3z = -8 + 5$
$-3z = -3$
$z = 1$
Substitute $z = 1$ : Substitute $z = 1$ into $x = z + 5$ . $\newline$ $x = 1 + 5$ $\newline$ $x = 6$
Check solution: We already have $y = 1$ from the second equation.
Check solution: Check the solution $(x=6, y=1, z=1)$ in the original equations. $\newline$ First equation: $-x + y - 2z = -7$ $\newline$ $-6 + 1 - 2(1) = -7$ $\newline$ $-6 + 1 - 2 = -7$ $\newline$ $-7 = -7$ (Checks out)
Check solution: Second equation: $y = 1$ $\newline$ $1 = 1$ (Checks out)
Check solution: Second equation: $y = 1$ $\newline$ $1 = 1$ (Checks out)Third equation: $x - 2y - z = 3$ $\newline$ $6 - 2(1) - 1 = 3$ $\newline$ $6 - 2 - 1 = 3$ $\newline$ $3 = 3$ (Checks out)

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