Solve the system of equations by elimination. $\newline$ $3x - y + 2z = -1$ $\newline$ $3x + y + 3z = 20$ $\newline$ $-x + y - 3z = -6$

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Algebra 2

Solve a system of equations in three variables using elimination

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Q. Solve the system of equations by elimination.

\newline

3x - y + 2z = -1

\newline

3x + y + 3z = 20

\newline

-x + y - 3z = -6

Eliminate $y$ by adding equations: Add the first and second equations to eliminate $y$ . $(3x - y + 2z) + (3x + y + 3z) = -1 + 20$ $6x + 5z = 19$
Find $x$ : Add the second and third equations to eliminate $y$ .
$(3x + y + 3z) + (-x + y - 3z) = 20 + (-6)$
$2x = 14$
$x = 7$
Substitute $x=7$ to find $y$ and $z$ : Substitute $x = 7$ into the first equation to find $y$ and $z$ . $\newline$ $3(7) - y + 2z = -1$ $\newline$ $21 - y + 2z = -1$ $\newline$ $-y + 2z = -22$
Substitute $x=7$ to find $y$ and $z$ : Substitute $x = 7$ into the second equation to find $y$ and $z$ . $\newline$ $3(7) + y + 3z = 20$ $\newline$ $21 + y + 3z = 20$ $\newline$ $y + 3z = -1$
Eliminate $z$ by subtracting equations: Subtract the modified first equation from the modified second equation to eliminate $z$ . $\$y + 3z$ - (-y + $2$ z) = $-1$ - ( $-22$ )\) $2y + z = 21$
Substitute $x=7$ to find $y$ and $z$ : Substitute $x = 7$ into the third equation to find $y$ and $z$ .
$-7 + y - 3z = -6$
$-7 + y - 3z = -6$
$y - 3z = 1$
Eliminate z by adding equations: Add the modified third equation to the equation $2y + z = 21$ to eliminate z. $\newline$ $(y - 3z) + (2y + z) = 1 + 21$ $\newline$ $3y - 2z = 22$
Eliminate z by adding equations: Now we have two equations with y and z: $\newline$ $-y + 2z = -22$ $\newline$ $3y - 2z = 22$ $\newline$ Add these two equations to eliminate z. $\newline$ $(-y + 2z) + (3y - 2z) = -22 + 22$ $\newline$ $2y = 0$ $\newline$ $y = 0$
Find $y$ : Substitute $y = 0$ into $y - 3z = 1$ to find $z$ .
$0 - 3z = 1$
$-3z = 1$
$z = -\frac{1}{3}$