Solve the System of Equations x+y-z=6 3x-2y+z=-5 x+3y-2z=14

$x + y - z = 6$ $3 x - 2 y + z = - 5$ $x + 3 y - 2 z = 14$

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

Subtract $y$ from both sides of the equation.

$x - z = 6 - y$

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

$x + 3 y - 2 z = 14$

Add $z$ to both sides of the equation.

$x = 6 - y + z$

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

$x + 3 y - 2 z = 14$

$x = 6 - y + z$

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

$x + 3 y - 2 z = 14$

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

Replace all occurrences of $x$ in $3 x - 2 y + z = - 5$ with $6 - y + z$ .

$3 (6 - y + z) - 2 y + z = - 5$

$x = 6 - y + z$

$x + 3 y - 2 z = 14$

Simplify the left side.

Simplify $3 (6 - y + z) - 2 y + z$ .

Simplify each term.

Apply the distributive property.

$3 \cdot 6 + 3 (- y) + 3 z - 2 y + z = - 5$

$x = 6 - y + z$

$x + 3 y - 2 z = 14$

Simplify.

Multiply $3$ by $6$ .

$18 + 3 (- y) + 3 z - 2 y + z = - 5$

$x = 6 - y + z$

$x + 3 y - 2 z = 14$

Multiply $- 1$ by $3$ .

$18 - 3 y + 3 z - 2 y + z = - 5$

$x = 6 - y + z$

$x + 3 y - 2 z = 14$

$18 - 3 y + 3 z - 2 y + z = - 5$

$x = 6 - y + z$

$x + 3 y - 2 z = 14$

$18 - 3 y + 3 z - 2 y + z = - 5$

$x = 6 - y + z$

$x + 3 y - 2 z = 14$

Simplify by adding terms.

Subtract $2 y$ from $- 3 y$ .

$18 - 5 y + 3 z + z = - 5$

$x = 6 - y + z$

$x + 3 y - 2 z = 14$

Add $3 z$ and $z$ .

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

$x + 3 y - 2 z = 14$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

$x + 3 y - 2 z = 14$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

$x + 3 y - 2 z = 14$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

$x + 3 y - 2 z = 14$

Replace all occurrences of $x$ in $x + 3 y - 2 z = 14$ with $6 - y + z$ .

$(6 - y + z) + 3 y - 2 z = 14$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

Simplify the left side.

Simplify $(6 - y + z) + 3 y - 2 z$ .

Add $- y$ and $3 y$ .

$6 + 2 y + z - 2 z = 14$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

Subtract $2 z$ from $z$ .

$6 + 2 y - z = 14$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

$6 + 2 y - z = 14$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

$6 + 2 y - z = 14$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

$6 + 2 y - z = 14$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

Solve for $y$ in $6 + 2 y - z = 14$ .

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

Subtract $6$ from both sides of the equation.

$2 y - z = 14 - 6$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

Add $z$ to both sides of the equation.

$2 y = 14 - 6 + z$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

Subtract $6$ from $14$ .

$2 y = 8 + z$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

$2 y = 8 + z$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

Divide each term in $2 y = 8 + z$ by $2$ and simplify.

Divide each term in $2 y = 8 + z$ by $2$ .

$\frac{2 y}{2} = \frac{8}{2} + \frac{z}{2}$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{8}{2} + \frac{z}{2}$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

Divide $y$ by $1$ .

$y = \frac{8}{2} + \frac{z}{2}$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

$y = \frac{8}{2} + \frac{z}{2}$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

$y = \frac{8}{2} + \frac{z}{2}$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

Simplify the right side.

Divide $8$ by $2$ .

$y = 4 + \frac{z}{2}$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

$y = 4 + \frac{z}{2}$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

$y = 4 + \frac{z}{2}$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

$y = 4 + \frac{z}{2}$

$18 - 5 y + 4 z = - 5$

$x = 6 - y + z$

Replace all occurrences of $y$ with $4 + \frac{z}{2}$ in each equation.

Replace all occurrences of $y$ in $18 - 5 y + 4 z = - 5$ with $4 + \frac{z}{2}$ .

$18 - 5 (4 + \frac{z}{2}) + 4 z = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Simplify the left side.

Simplify $18 - 5 (4 + \frac{z}{2}) + 4 z$ .

Simplify each term.

Apply the distributive property.

$18 - 5 \cdot 4 - 5 \frac{z}{2} + 4 z = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Multiply $- 5$ by $4$ .

$18 - 20 - 5 \frac{z}{2} + 4 z = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Combine $- 5$ and $\frac{z}{2}$ .

$18 - 20 + \frac{- 5 z}{2} + 4 z = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Move the negative in front of the fraction.

$18 - 20 - \frac{5 z}{2} + 4 z = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

$18 - 20 - \frac{5 z}{2} + 4 z = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Subtract $20$ from $18$ .

$- 2 - \frac{5 z}{2} + 4 z = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

To write $4 z$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- 2 - \frac{5 z}{2} + 4 z \cdot \frac{2}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Simplify terms.

Combine $4 z$ and $\frac{2}{2}$ .

$- 2 - \frac{5 z}{2} + \frac{4 z \cdot 2}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Combine the numerators over the common denominator.

$- 2 + \frac{- 5 z + 4 z \cdot 2}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

$- 2 + \frac{- 5 z + 4 z \cdot 2}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Simplify each term.

Simplify the numerator.

Factor $z$ out of $- 5 z + 4 z \cdot 2$ .

Factor $z$ out of $- 5 z$ .

$- 2 + \frac{z \cdot - 5 + 4 z \cdot 2}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Factor $z$ out of $4 z \cdot 2$ .

$- 2 + \frac{z \cdot - 5 + z (4 \cdot 2)}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Factor $z$ out of $z \cdot - 5 + z (4 \cdot 2)$ .

$- 2 + \frac{z (- 5 + 4 \cdot 2)}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

$- 2 + \frac{z (- 5 + 4 \cdot 2)}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Multiply $4$ by $2$ .

$- 2 + \frac{z (- 5 + 8)}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Add $- 5$ and $8$ .

$- 2 + \frac{z \cdot 3}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

$- 2 + \frac{z \cdot 3}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Move $3$ to the left of $z$ .

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - y + z$

Replace all occurrences of $y$ in $x = 6 - y + z$ with $4 + \frac{z}{2}$ .

$x = 6 - (4 + \frac{z}{2}) + z$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

Simplify the right side.

Simplify $6 - (4 + \frac{z}{2}) + z$ .

Simplify each term.

Apply the distributive property.

$x = 6 - 1 \cdot 4 - \frac{z}{2} + z$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

Multiply $- 1$ by $4$ .

$x = 6 - 4 - \frac{z}{2} + z$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 6 - 4 - \frac{z}{2} + z$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

Subtract $4$ from $6$ .

$x = 2 - \frac{z}{2} + z$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

To write $z$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = 2 - \frac{z}{2} + z \cdot \frac{2}{2}$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

Combine $z$ and $\frac{2}{2}$ .

$x = 2 - \frac{z}{2} + \frac{z \cdot 2}{2}$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

Combine the numerators over the common denominator.

$x = 2 + \frac{- z + z \cdot 2}{2}$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

Add $- z$ and $z \cdot 2$ .

Reorder $z$ and $2$ .

$x = 2 + \frac{- z + 2 \cdot z}{2}$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

Add $- z$ and $2 \cdot z$ .

$x = 2 + \frac{z}{2}$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 2 + \frac{z}{2}$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 2 + \frac{z}{2}$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 2 + \frac{z}{2}$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

$x = 2 + \frac{z}{2}$

$- 2 + \frac{3 z}{2} = - 5$

$y = 4 + \frac{z}{2}$

Solve for $z$ in $- 2 + \frac{3 z}{2} = - 5$ .

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

Add $2$ to both sides of the equation.

$\frac{3 z}{2} = - 5 + 2$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

Add $- 5$ and $2$ .

$\frac{3 z}{2} = - 3$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

$\frac{3 z}{2} = - 3$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

Multiply both sides of the equation by $\frac{2}{3}$ .

$\frac{2}{3} \cdot \frac{3 z}{2} = \frac{2}{3} \cdot - 3$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{2}{3} \cdot \frac{3 z}{2}$ .

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2}{3} \cdot \frac{3 z}{2} = \frac{2}{3} \cdot - 3$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

Rewrite the expression.

$\frac{1}{3} \cdot (3 z) = \frac{2}{3} \cdot - 3$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

$\frac{1}{3} \cdot (3 z) = \frac{2}{3} \cdot - 3$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

Cancel the common factor of $3$ .

Factor $3$ out of $3 z$ .

$\frac{1}{3} \cdot (3 (z)) = \frac{2}{3} \cdot - 3$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

Cancel the common factor.

$\frac{1}{3} \cdot (3 z) = \frac{2}{3} \cdot - 3$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

Rewrite the expression.

$z = \frac{2}{3} \cdot - 3$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

$z = \frac{2}{3} \cdot - 3$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

$z = \frac{2}{3} \cdot - 3$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

$z = \frac{2}{3} \cdot - 3$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

Simplify the right side.

Simplify $\frac{2}{3} \cdot - 3$ .

Cancel the common factor of $3$ .

Factor $3$ out of $- 3$ .

$z = \frac{2}{3} \cdot (3 (- 1))$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

Cancel the common factor.

$z = \frac{2}{3} \cdot (3 \cdot - 1)$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

Rewrite the expression.

$z = 2 \cdot - 1$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

$z = 2 \cdot - 1$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

Multiply $2$ by $- 1$ .

$z = - 2$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

$z = - 2$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

$z = - 2$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

$z = - 2$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

$z = - 2$

$x = 2 + \frac{z}{2}$

$y = 4 + \frac{z}{2}$

Replace all occurrences of $z$ with $- 2$ in each equation.

Replace all occurrences of $z$ in $x = 2 + \frac{z}{2}$ with $- 2$ .

$x = 2 + \frac{- 2}{2}$

$z = - 2$

$y = 4 + \frac{z}{2}$

Simplify the right side.

Simplify $2 + \frac{- 2}{2}$ .

Divide $- 2$ by $2$ .

$x = 2 - 1$

$z = - 2$

$y = 4 + \frac{z}{2}$

Subtract $1$ from $2$ .

$x = 1$

$z = - 2$

$y = 4 + \frac{z}{2}$

$x = 1$

$z = - 2$

$y = 4 + \frac{z}{2}$

$x = 1$

$z = - 2$

$y = 4 + \frac{z}{2}$

Replace all occurrences of $z$ in $y = 4 + \frac{z}{2}$ with $- 2$ .

$y = 4 + \frac{- 2}{2}$

$x = 1$

$z = - 2$

Simplify the right side.

Simplify $4 + \frac{- 2}{2}$ .

Divide $- 2$ by $2$ .

$y = 4 - 1$

$x = 1$

$z = - 2$

Subtract $1$ from $4$ .

$y = 3$

$x = 1$

$z = - 2$

$y = 3$

$x = 1$

$z = - 2$

$y = 3$

$x = 1$

$z = - 2$

$y = 3$

$x = 1$

$z = - 2$

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

$(1, 3, - 2)$

The result can be shown in multiple forms.

Point Form:

$(1, 3, - 2)$

Equation Form:

$x = 1, y = 3, z = - 2$

Solve the System of Equations x+y-z=6 3x-2y+z=-5 x+3y-2z=14

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