Solve the system of equations by substitution. x - 3y - z = 16 -2x + 3y - 2z = -17 z = 3 ____

Substitute z = 3: Substitute z = 3 into the first equation x - 3y - z = 16. x - 3y - 3 = 16 x - 3y = 19Substitute z = 3: Substitute z = 3 into the second equation -2x + 3y - 2z = -17. -2x + 3y - 2*3 = -17 -2x + 3y - 6 = -17 -2x + 3y = -11Two Equations with Two Variables: Now we have two equations with two variables: 1) x - 3y = 19 2) -2x + 3y = -11 Add the two equations to eliminate y. (x - 3y) + (-2x + 3y) = 19 + (-11) x - 2x = 8 -x = 8 x = -8

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Solve the system of equations by substitution. $\newline$ $x - 3y - z = 16$ $\newline$ $-2x + 3y - 2z = -17$ $\newline$ $z = 3$

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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 - 3y - z = 16

\newline

-2x + 3y - 2z = -17

\newline

z = 3

Substitute $z = 3$ : Substitute $z = 3$ into the first equation $x - 3y - z = 16$ . $\newline$ $x - 3y - 3 = 16$ $\newline$ $x - 3y = 19$
Substitute $z = 3$ : Substitute $z = 3$ into the second equation $-2x + 3y - 2z = -17$ . $\newline$ $-2x + 3y - 2\times3 = -17$ $\newline$ $-2x + 3y - 6 = -17$ $\newline$ $-2x + 3y = -11$
Two Equations with Two Variables: Now we have two equations with two variables: $\newline$ $1$ ) $x - 3y = 19$ $\newline$ $2$ ) $-2x + 3y = -11$ $\newline$ Add the two equations to eliminate $y$ . $\newline$ $(x - 3y) + (-2x + 3y) = 19 + (-11)$ $\newline$ $x - 2x = 8$ $\newline$ $-x = 8$ $\newline$ $x = -8$