Solve the System of Equations x+3y=1 -2y-z=1 4x-3z=13

$x + 3 y = 1$ $- 2 y - z = 1$ $4 x - 3 z = 13$

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

$x = 1 - 3 y$

$- 2 y - z = 1$

$4 x - 3 z = 13$

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

Replace all occurrences of $x$ in $4 x - 3 z = 13$ with $1 - 3 y$ .

$4 (1 - 3 y) - 3 z = 13$

$x = 1 - 3 y$

$- 2 y - z = 1$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$4 \cdot 1 + 4 (- 3 y) - 3 z = 13$

$x = 1 - 3 y$

$- 2 y - z = 1$

Multiply $4$ by $1$ .

$4 + 4 (- 3 y) - 3 z = 13$

$x = 1 - 3 y$

$- 2 y - z = 1$

Multiply $- 3$ by $4$ .

$4 - 12 y - 3 z = 13$

$x = 1 - 3 y$

$- 2 y - z = 1$

$4 - 12 y - 3 z = 13$

$x = 1 - 3 y$

$- 2 y - z = 1$

$4 - 12 y - 3 z = 13$

$x = 1 - 3 y$

$- 2 y - z = 1$

$4 - 12 y - 3 z = 13$

$x = 1 - 3 y$

$- 2 y - z = 1$

Reorder $1$ and $- 3 y$ .

$x = - 3 y + 1$

$4 - 12 y - 3 z = 13$

$- 2 y - z = 1$

Solve for $y$ in $4 - 12 y - 3 z = 13$ .

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

Subtract $4$ from both sides of the equation.

$- 12 y - 3 z = 13 - 4$

$x = - 3 y + 1$

$- 2 y - z = 1$

Add $3 z$ to both sides of the equation.

$- 12 y = 13 - 4 + 3 z$

$x = - 3 y + 1$

$- 2 y - z = 1$

Subtract $4$ from $13$ .

$- 12 y = 9 + 3 z$

$x = - 3 y + 1$

$- 2 y - z = 1$

$- 12 y = 9 + 3 z$

$x = - 3 y + 1$

$- 2 y - z = 1$

Divide each term in $- 12 y = 9 + 3 z$ by $- 12$ and simplify.

Divide each term in $- 12 y = 9 + 3 z$ by $- 12$ .

$\frac{- 12 y}{- 12} = \frac{9}{- 12} + \frac{3 z}{- 12}$

$x = - 3 y + 1$

$- 2 y - z = 1$

Simplify the left side.

Cancel the common factor of $- 12$ .

Cancel the common factor.

$\frac{- 12 y}{- 12} = \frac{9}{- 12} + \frac{3 z}{- 12}$

$x = - 3 y + 1$

$- 2 y - z = 1$

Divide $y$ by $1$ .

$y = \frac{9}{- 12} + \frac{3 z}{- 12}$

$x = - 3 y + 1$

$- 2 y - z = 1$

$y = \frac{9}{- 12} + \frac{3 z}{- 12}$

$x = - 3 y + 1$

$- 2 y - z = 1$

$y = \frac{9}{- 12} + \frac{3 z}{- 12}$

$x = - 3 y + 1$

$- 2 y - z = 1$

Simplify the right side.

Simplify each term.

Cancel the common factor of $9$ and $- 12$ .

Factor $3$ out of $9$ .

$y = \frac{3 (3)}{- 12} + \frac{3 z}{- 12}$

$x = - 3 y + 1$

$- 2 y - z = 1$

Cancel the common factors.

Factor $3$ out of $- 12$ .

$y = \frac{3 \cdot 3}{3 \cdot - 4} + \frac{3 z}{- 12}$

$x = - 3 y + 1$

$- 2 y - z = 1$

Cancel the common factor.

$y = \frac{3 \cdot 3}{3 \cdot - 4} + \frac{3 z}{- 12}$

$x = - 3 y + 1$

$- 2 y - z = 1$

Rewrite the expression.

$y = \frac{3}{- 4} + \frac{3 z}{- 12}$

$x = - 3 y + 1$

$- 2 y - z = 1$

$y = \frac{3}{- 4} + \frac{3 z}{- 12}$

$x = - 3 y + 1$

$- 2 y - z = 1$

$y = \frac{3}{- 4} + \frac{3 z}{- 12}$

$x = - 3 y + 1$

$- 2 y - z = 1$

Move the negative in front of the fraction.

$y = - \frac{3}{4} + \frac{3 z}{- 12}$

$x = - 3 y + 1$

$- 2 y - z = 1$

Cancel the common factor of $3$ and $- 12$ .

Factor $3$ out of $3 z$ .

$y = - \frac{3}{4} + \frac{3 (z)}{- 12}$

$x = - 3 y + 1$

$- 2 y - z = 1$

Cancel the common factors.

Factor $3$ out of $- 12$ .

$y = - \frac{3}{4} + \frac{3 z}{3 \cdot - 4}$

$x = - 3 y + 1$

$- 2 y - z = 1$

Cancel the common factor.

$y = - \frac{3}{4} + \frac{3 z}{3 \cdot - 4}$

$x = - 3 y + 1$

$- 2 y - z = 1$

Rewrite the expression.

$y = - \frac{3}{4} + \frac{z}{- 4}$

$x = - 3 y + 1$

$- 2 y - z = 1$

$y = - \frac{3}{4} + \frac{z}{- 4}$

$x = - 3 y + 1$

$- 2 y - z = 1$

$y = - \frac{3}{4} + \frac{z}{- 4}$

$x = - 3 y + 1$

$- 2 y - z = 1$

Move the negative in front of the fraction.

$y = - \frac{3}{4} - \frac{z}{4}$

$x = - 3 y + 1$

$- 2 y - z = 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$x = - 3 y + 1$

$- 2 y - z = 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$x = - 3 y + 1$

$- 2 y - z = 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$x = - 3 y + 1$

$- 2 y - z = 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$x = - 3 y + 1$

$- 2 y - z = 1$

Replace all occurrences of $y$ with $- \frac{3}{4} - \frac{z}{4}$ in each equation.

Replace all occurrences of $y$ in $x = - 3 y + 1$ with $- \frac{3}{4} - \frac{z}{4}$ .

$x = - 3 (- \frac{3}{4} - \frac{z}{4}) + 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

Simplify the right side.

Simplify $- 3 (- \frac{3}{4} - \frac{z}{4}) + 1$ .

Simplify each term.

Apply the distributive property.

$x = - 3 (- \frac{3}{4}) - 3 (- \frac{z}{4}) + 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

Multiply $- 3 (- \frac{3}{4})$ .

Multiply $- 1$ by $- 3$ .

$x = 3 (\frac{3}{4}) - 3 (- \frac{z}{4}) + 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

Combine $3$ and $\frac{3}{4}$ .

$x = \frac{3 \cdot 3}{4} - 3 (- \frac{z}{4}) + 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

Multiply $3$ by $3$ .

$x = \frac{9}{4} - 3 (- \frac{z}{4}) + 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

$x = \frac{9}{4} - 3 (- \frac{z}{4}) + 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

Multiply $- 3 (- \frac{z}{4})$ .

Multiply $- 1$ by $- 3$ .

$x = \frac{9}{4} + 3 (\frac{z}{4}) + 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

Combine $3$ and $\frac{z}{4}$ .

$x = \frac{9}{4} + \frac{3 z}{4} + 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

$x = \frac{9}{4} + \frac{3 z}{4} + 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

$x = \frac{9}{4} + \frac{3 z}{4} + 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

Simplify the expression.

Write $1$ as a fraction with a common denominator.

$x = \frac{3 z}{4} + \frac{9}{4} + \frac{4}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

Combine the numerators over the common denominator.

$x = \frac{3 z}{4} + \frac{9 + 4}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

Add $9$ and $4$ .

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 2 y - z = 1$

Replace all occurrences of $y$ in $- 2 y - z = 1$ with $- \frac{3}{4} - \frac{z}{4}$ .

$- 2 (- \frac{3}{4} - \frac{z}{4}) - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Simplify the left side.

Simplify $- 2 (- \frac{3}{4} - \frac{z}{4}) - z$ .

Simplify each term.

Apply the distributive property.

$- 2 (- \frac{3}{4}) - 2 (- \frac{z}{4}) - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3}{4}$ into the numerator.

$- 2 (\frac{- 3}{4}) - 2 (- \frac{z}{4}) - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Factor $2$ out of $- 2$ .

$2 (- 1) (\frac{- 3}{4}) - 2 (- \frac{z}{4}) - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Factor $2$ out of $4$ .

$2 \cdot (- 1 \frac{- 3}{2 \cdot 2}) - 2 (- \frac{z}{4}) - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Cancel the common factor.

$2 \cdot (- 1 \frac{- 3}{2 \cdot 2}) - 2 (- \frac{z}{4}) - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Rewrite the expression.

$- 1 (\frac{- 3}{2}) - 2 (- \frac{z}{4}) - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$- 1 (\frac{- 3}{2}) - 2 (- \frac{z}{4}) - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{z}{4}$ into the numerator.

$- 1 (\frac{- 3}{2}) - 2 \frac{- z}{4} - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Factor $2$ out of $- 2$ .

$- 1 (\frac{- 3}{2}) + 2 (- 1) (\frac{- z}{4}) - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Factor $2$ out of $4$ .

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Cancel the common factor.

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Rewrite the expression.

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Simplify each term.

Move the negative in front of the fraction.

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Multiply $- 1 (- \frac{3}{2})$ .

Multiply $- 1$ by $- 1$ .

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Multiply $\frac{3}{2}$ by $1$ .

$\frac{3}{2} - 1 \frac{- z}{2} - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$\frac{3}{2} - 1 \frac{- z}{2} - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Move the negative in front of the fraction.

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Multiply $- 1 (- \frac{z}{2})$ .

Multiply $- 1$ by $- 1$ .

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Multiply $\frac{z}{2}$ by $1$ .

$\frac{3}{2} + \frac{z}{2} - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$\frac{3}{2} + \frac{z}{2} - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$\frac{3}{2} + \frac{z}{2} - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$\frac{3}{2} + \frac{z}{2} - z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

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

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

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

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Combine the numerators over the common denominator.

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Combine the numerators over the common denominator.

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Multiply $2$ by $- 1$ .

$\frac{3 + z - 2 z}{2} = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Subtract $2 z$ from $z$ .

$\frac{3 - z}{2} = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$\frac{3 - z}{2} = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$\frac{3 - z}{2} = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$\frac{3 - z}{2} = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Solve for $z$ in $\frac{3 - z}{2} = 1$ .

Multiply both sides by $2$ .

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Simplify.

Simplify the left side.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Rewrite the expression.

$3 - z = 1 \cdot 2$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$3 - z = 1 \cdot 2$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Reorder $3$ and $- z$ .

$- z + 3 = 1 \cdot 2$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$- z + 3 = 1 \cdot 2$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$- z + 3 = 1 \cdot 2$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Simplify the right side.

Multiply $2$ by $1$ .

$- z + 3 = 2$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$- z + 3 = 2$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$- z + 3 = 2$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Solve for $z$ .

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

Subtract $3$ from both sides of the equation.

$- z = 2 - 3$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Subtract $3$ from $2$ .

$- z = - 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$- z = - 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Divide each term in $- z = - 1$ by $- 1$ and simplify.

Divide each term in $- z = - 1$ by $- 1$ .

$\frac{- z}{- 1} = \frac{- 1}{- 1}$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{z}{1} = \frac{- 1}{- 1}$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Divide $z$ by $1$ .

$z = \frac{- 1}{- 1}$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$z = \frac{- 1}{- 1}$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

$z = 1$

$x = \frac{3 z}{4} + \frac{13}{4}$

$y = - \frac{3}{4} - \frac{z}{4}$

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

Replace all occurrences of $z$ in $x = \frac{3 z}{4} + \frac{13}{4}$ with $1$ .

$x = \frac{3 (1)}{4} + \frac{13}{4}$

$z = 1$

$y = - \frac{3}{4} - \frac{z}{4}$

Simplify the right side.

Simplify $\frac{3 (1)}{4} + \frac{13}{4}$ .

Combine the numerators over the common denominator.

$x = \frac{3 (1) + 13}{4}$

$z = 1$

$y = - \frac{3}{4} - \frac{z}{4}$

Simplify the expression.

Multiply $3$ by $1$ .

$x = \frac{3 + 13}{4}$

$z = 1$

$y = - \frac{3}{4} - \frac{z}{4}$

Add $3$ and $13$ .

$x = \frac{16}{4}$

$z = 1$

$y = - \frac{3}{4} - \frac{z}{4}$

Divide $16$ by $4$ .

$x = 4$

$z = 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$x = 4$

$z = 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$x = 4$

$z = 1$

$y = - \frac{3}{4} - \frac{z}{4}$

$x = 4$

$z = 1$

$y = - \frac{3}{4} - \frac{z}{4}$

Replace all occurrences of $z$ in $y = - \frac{3}{4} - \frac{z}{4}$ with $1$ .

$y = - \frac{3}{4} - \frac{1}{4}$

$x = 4$

$z = 1$

Simplify the right side.

Simplify $- \frac{3}{4} - \frac{1}{4}$ .

Combine the numerators over the common denominator.

$y = \frac{- 3 - 1}{4}$

$x = 4$

$z = 1$

Simplify the expression.

Subtract $1$ from $- 3$ .

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

$x = 4$

$z = 1$

Divide $- 4$ by $4$ .

$y = - 1$

$x = 4$

$z = 1$

$y = - 1$

$x = 4$

$z = 1$

$y = - 1$

$x = 4$

$z = 1$

$y = - 1$

$x = 4$

$z = 1$

$y = - 1$

$x = 4$

$z = 1$

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

$(4, - 1, 1)$

The result can be shown in multiple forms.

Point Form:

$(4, - 1, 1)$

Equation Form:

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

Solve the System of Equations x+3y=1 -2y-z=1 4x-3z=13

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