Solve the system of equations by substitution. x + y + 2z = -1 y = -6 -2x + y + 2z = -4 ____

Substitute y = -6: Substitute y = -6 into the first equation x + y + 2z = -1. x + (-6) + 2z = -1 x - 6 + 2z = -1Add 6 to isolate x: Add 6 to both sides to isolate x + 2z. x + 2z = -1 + 6 x + 2z = 5Substitute y = -6: Substitute y = -6 into the third equation -2x + y + 2z = -4. -2x + (-6) + 2z = -4 -2x - 6 + 2z = -4Add 6 to isolate -2x: Add 6 to both sides to isolate -2x + 2z. -2x + 2z = -4 + 6 -2x + 2z = 2Divide by -2 to solve x: Divide the entire equation by -2 to solve for x. x = 2 / -2 x = -1

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Solve the system of equations by substitution. $\newline$ $x + y + 2z = -1$ $\newline$ $y = -6$ $\newline$ $-2x + y + 2z = -4$

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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

x + y + 2z = -1

\newline

y = -6

\newline

-2x + y + 2z = -4

Substitute $y = -6$ : Substitute $y = -6$ into the first equation $x + y + 2z = -1$ . $\newline$ $x + (-6) + 2z = -1$ $\newline$ $x - 6 + 2z = -1$
Add $6$ to isolate x: Add $6$ to both sides to isolate $x + 2z$ . $\newline$ $x + 2z = -1 + 6$ $\newline$ $x + 2z = 5$
Substitute $y = -6$ : Substitute $y = -6$ into the third equation $-2x + y + 2z = -4$ . $\newline$ $-2x + (-6) + 2z = -4$ $\newline$ $-2x - 6 + 2z = -4$
Add $6$ to isolate $-2x$ : Add $6$ to both sides to isolate $-2x + 2z$ . $\newline$ $-2x + 2z = -4 + 6$ $\newline$ $-2x + 2z = 2$
Divide by $-2$ to solve $x$ : Divide the entire equation by $-2$ to solve for $x$ .
$x = \frac{2}{-2}$
$x = -1$