Solve the System of Equations x+3z=-1 y-2z=6 4x+3y=20

$x + 3 z = - 1$ $y - 2 z = 6$ $4 x + 3 y = 20$

Step 1

Subtract $3 z$ from both sides of the equation.

$x = - 1 - 3 z$

$y - 2 z = 6$

$4 x + 3 y = 20$

Step 2

Replace all occurrences of $x$ with $- 1 - 3 z$ in each equation.

Replace all occurrences of $x$ in $4 x + 3 y = 20$ with $- 1 - 3 z$ .

$4 (- 1 - 3 z) + 3 y = 20$

$x = - 1 - 3 z$

$y - 2 z = 6$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$4 \cdot - 1 + 4 (- 3 z) + 3 y = 20$

$x = - 1 - 3 z$

$y - 2 z = 6$

Multiply $4$ by $- 1$ .

$- 4 + 4 (- 3 z) + 3 y = 20$

$x = - 1 - 3 z$

$y - 2 z = 6$

Multiply $- 3$ by $4$ .

$- 4 - 12 z + 3 y = 20$

$x = - 1 - 3 z$

$y - 2 z = 6$

$- 4 - 12 z + 3 y = 20$

$x = - 1 - 3 z$

$y - 2 z = 6$

$- 4 - 12 z + 3 y = 20$

$x = - 1 - 3 z$

$y - 2 z = 6$

$- 4 - 12 z + 3 y = 20$

$x = - 1 - 3 z$

$y - 2 z = 6$

Step 3

Reorder $- 1$ and $- 3 z$ .

$x = - 3 z - 1$

$- 4 - 12 z + 3 y = 20$

$y - 2 z = 6$

Step 4

Add $2 z$ to both sides of the equation.

$y = 6 + 2 z$

$x = - 3 z - 1$

$- 4 - 12 z + 3 y = 20$

Step 5

Replace all occurrences of $y$ with $6 + 2 z$ in each equation.

Replace all occurrences of $y$ in $- 4 - 12 z + 3 y = 20$ with $6 + 2 z$ .

$- 4 - 12 z + 3 (6 + 2 z) = 20$

$y = 6 + 2 z$

$x = - 3 z - 1$

Simplify the left side.

Simplify $- 4 - 12 z + 3 (6 + 2 z)$ .

Simplify each term.

Apply the distributive property.

$- 4 - 12 z + 3 \cdot 6 + 3 (2 z) = 20$

$y = 6 + 2 z$

$x = - 3 z - 1$

Multiply $3$ by $6$ .

$- 4 - 12 z + 18 + 3 (2 z) = 20$

$y = 6 + 2 z$

$x = - 3 z - 1$

Multiply $2$ by $3$ .

$- 4 - 12 z + 18 + 6 z = 20$

$y = 6 + 2 z$

$x = - 3 z - 1$

$- 4 - 12 z + 18 + 6 z = 20$

$y = 6 + 2 z$

$x = - 3 z - 1$

Simplify by adding terms.

Add $- 4$ and $18$ .

$- 12 z + 14 + 6 z = 20$

$y = 6 + 2 z$

$x = - 3 z - 1$

Add $- 12 z$ and $6 z$ .

$- 6 z + 14 = 20$

$y = 6 + 2 z$

$x = - 3 z - 1$

$- 6 z + 14 = 20$

$y = 6 + 2 z$

$x = - 3 z - 1$

$- 6 z + 14 = 20$

$y = 6 + 2 z$

$x = - 3 z - 1$

$- 6 z + 14 = 20$

$y = 6 + 2 z$

$x = - 3 z - 1$

$- 6 z + 14 = 20$

$y = 6 + 2 z$

$x = - 3 z - 1$

Step 6

Solve for $z$ in $- 6 z + 14 = 20$ .

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

Subtract $14$ from both sides of the equation.

$- 6 z = 20 - 14$

$y = 6 + 2 z$

$x = - 3 z - 1$

Subtract $14$ from $20$ .

$- 6 z = 6$

$y = 6 + 2 z$

$x = - 3 z - 1$

$- 6 z = 6$

$y = 6 + 2 z$

$x = - 3 z - 1$

Divide each term in $- 6 z = 6$ by $- 6$ and simplify.

Divide each term in $- 6 z = 6$ by $- 6$ .

$\frac{- 6 z}{- 6} = \frac{6}{- 6}$

$y = 6 + 2 z$

$x = - 3 z - 1$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 z}{- 6} = \frac{6}{- 6}$

$y = 6 + 2 z$

$x = - 3 z - 1$

Divide $z$ by $1$ .

$z = \frac{6}{- 6}$

$y = 6 + 2 z$

$x = - 3 z - 1$

$z = \frac{6}{- 6}$

$y = 6 + 2 z$

$x = - 3 z - 1$

$z = \frac{6}{- 6}$

$y = 6 + 2 z$

$x = - 3 z - 1$

Simplify the right side.

Divide $6$ by $- 6$ .

$z = - 1$

$y = 6 + 2 z$

$x = - 3 z - 1$

$z = - 1$

$y = 6 + 2 z$

$x = - 3 z - 1$

$z = - 1$

$y = 6 + 2 z$

$x = - 3 z - 1$

$z = - 1$

$y = 6 + 2 z$

$x = - 3 z - 1$

Step 7

Replace all occurrences of $z$ with $- 1$ in each equation.

Replace all occurrences of $z$ in $y = 6 + 2 z$ with $- 1$ .

$y = 6 + 2 (- 1)$

$z = - 1$

$x = - 3 z - 1$

Simplify the right side.

Simplify $6 + 2 (- 1)$ .

Multiply $2$ by $- 1$ .

$y = 6 - 2$

$z = - 1$

$x = - 3 z - 1$

Subtract $2$ from $6$ .

$y = 4$

$z = - 1$

$x = - 3 z - 1$

$y = 4$

$z = - 1$

$x = - 3 z - 1$

$y = 4$

$z = - 1$

$x = - 3 z - 1$

Replace all occurrences of $z$ in $x = - 3 z - 1$ with $- 1$ .

$x = - 3 \cdot - 1 - 1$

$y = 4$

$z = - 1$

Simplify the right side.

Simplify $- 3 \cdot - 1 - 1$ .

Multiply $- 3$ by $- 1$ .

$x = 3 - 1$

$y = 4$

$z = - 1$

Subtract $1$ from $3$ .

$x = 2$

$y = 4$

$z = - 1$

$x = 2$

$y = 4$

$z = - 1$

$x = 2$

$y = 4$

$z = - 1$

$x = 2$

$y = 4$

$z = - 1$

Step 8

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

$(2, 4, - 1)$

Step 9

The result can be shown in multiple forms.

Point Form:

$(2, 4, - 1)$

Equation Form:

$x = 2, y = 4, z = - 1$

Solve the System of Equations x+3z=-1 y-2z=6 4x+3y=20

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