Solve the System of Equations x+y-3z=-20 -3y+2z=5 z=4

$x + y - 3 z = - 20$ $- 3 y + 2 z = 5$ $z = 4$

Step 1

Replace all occurrences of $z$ with $4$ in each equation.

Replace all occurrences of $z$ in $x + y - 3 z = - 20$ with $4$ .

$x + y - 3 \cdot 4 = - 20$

$z = 4$

$- 3 y + 2 z = 5$

Simplify the left side.

Multiply $- 3$ by $4$ .

$x + y - 12 = - 20$

$z = 4$

$- 3 y + 2 z = 5$

$x + y - 12 = - 20$

$z = 4$

$- 3 y + 2 z = 5$

Replace all occurrences of $z$ in $- 3 y + 2 z = 5$ with $4$ .

$- 3 y + 2 (4) = 5$

$x + y - 12 = - 20$

$z = 4$

Simplify the left side.

Multiply $2$ by $4$ .

$- 3 y + 8 = 5$

$x + y - 12 = - 20$

$z = 4$

$- 3 y + 8 = 5$

$x + y - 12 = - 20$

$z = 4$

$- 3 y + 8 = 5$

$x + y - 12 = - 20$

$z = 4$

Step 2

Solve for $y$ in $- 3 y + 8 = 5$ .

Move all terms not containing $y$ to the right side of the equation.

Subtract $8$ from both sides of the equation.

$- 3 y = 5 - 8$

$x + y - 12 = - 20$

$z = 4$

Subtract $8$ from $5$ .

$- 3 y = - 3$

$x + y - 12 = - 20$

$z = 4$

$- 3 y = - 3$

$x + y - 12 = - 20$

$z = 4$

Divide each term in $- 3 y = - 3$ by $- 3$ and simplify.

Divide each term in $- 3 y = - 3$ by $- 3$ .

$\frac{- 3 y}{- 3} = \frac{- 3}{- 3}$

$x + y - 12 = - 20$

$z = 4$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 y}{- 3} = \frac{- 3}{- 3}$

$x + y - 12 = - 20$

$z = 4$

Divide $y$ by $1$ .

$y = \frac{- 3}{- 3}$

$x + y - 12 = - 20$

$z = 4$

$y = \frac{- 3}{- 3}$

$x + y - 12 = - 20$

$z = 4$

$y = \frac{- 3}{- 3}$

$x + y - 12 = - 20$

$z = 4$

Simplify the right side.

Divide $- 3$ by $- 3$ .

$y = 1$

$x + y - 12 = - 20$

$z = 4$

$y = 1$

$x + y - 12 = - 20$

$z = 4$

$y = 1$

$x + y - 12 = - 20$

$z = 4$

$y = 1$

$x + y - 12 = - 20$

$z = 4$

Step 3

Replace all occurrences of $y$ with $1$ in each equation.

Replace all occurrences of $y$ in $x + y - 12 = - 20$ with $1$ .

$x + 1 - 12 = - 20$

$y = 1$

$z = 4$

Simplify the left side.

Simplify $x + 1 - 12$ .

Remove parentheses.

$x + 1 - 12 = - 20$

$y = 1$

$z = 4$

Subtract $12$ from $1$ .

$x - 11 = - 20$

$y = 1$

$z = 4$

$x - 11 = - 20$

$y = 1$

$z = 4$

$x - 11 = - 20$

$y = 1$

$z = 4$

$x - 11 = - 20$

$y = 1$

$z = 4$

Step 4

Move all terms not containing $x$ to the right side of the equation.

Add $11$ to both sides of the equation.

$x = - 20 + 11$

$y = 1$

$z = 4$

Add $- 20$ and $11$ .

$x = - 9$

$y = 1$

$z = 4$

$x = - 9$

$y = 1$

$z = 4$

Step 5

Solve the system of equations $x = - 9 y = 1 z = 4$ .

$x = - 9 y = 1 z = 4$

Step 6

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(- 9, 1, 4)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(- 9, 1, 4)$

Equation Form:

$x = - 9, y = 1, z = 4$

Solve the System of Equations x+y-3z=-20 -3y+2z=5 z=4

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