Solve the System of Equations 3x-2y+z=5 2x+y+z=1 x+4y-2z=-3

$3 x - 2 y + z = 5$ $2 x + y + z = 1$ $x + 4 y - 2 z = - 3$

Step 1

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

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

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

$2 x + y + z = 1$

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

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

$z = 5 - 3 x + 2 y$

$2 x + y + z = 1$

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

$z = 5 - 3 x + 2 y$

$2 x + y + z = 1$

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

Step 2

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

Replace all occurrences of $z$ in $2 x + y + z = 1$ with $5 - 3 x + 2 y$ .

$2 x + y + 5 - 3 x + 2 y = 1$

$z = 5 - 3 x + 2 y$

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

Simplify the left side.

Simplify $2 x + y + 5 - 3 x + 2 y$ .

Remove parentheses.

$2 x + y + 5 - 3 x + 2 y = 1$

$z = 5 - 3 x + 2 y$

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

Subtract $3 x$ from $2 x$ .

$- x + y + 5 + 2 y = 1$

$z = 5 - 3 x + 2 y$

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

Add $y$ and $2 y$ .

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

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

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

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

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

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

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

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

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Simplify the left side.

Simplify $x + 4 y - 2 (5 - 3 x + 2 y)$ .

Simplify each term.

Apply the distributive property.

$x + 4 y - 2 \cdot 5 - 2 (- 3 x) - 2 (2 y) = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Simplify.

Multiply $- 2$ by $5$ .

$x + 4 y - 10 - 2 (- 3 x) - 2 (2 y) = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Multiply $- 3$ by $- 2$ .

$x + 4 y - 10 + 6 x - 2 (2 y) = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Multiply $2$ by $- 2$ .

$x + 4 y - 10 + 6 x - 4 y = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

$x + 4 y - 10 + 6 x - 4 y = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

$x + 4 y - 10 + 6 x - 4 y = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Simplify by adding terms.

Combine the opposite terms in $x + 4 y - 10 + 6 x - 4 y$ .

Subtract $4 y$ from $4 y$ .

$x - 10 + 6 x + 0 = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Add $x - 10 + 6 x$ and $0$ .

$x - 10 + 6 x = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

$x - 10 + 6 x = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Add $x$ and $6 x$ .

$7 x - 10 = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

$7 x - 10 = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

$7 x - 10 = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

$7 x - 10 = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

$7 x - 10 = - 3$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Step 3

Solve for $x$ in $7 x - 10 = - 3$ .

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

Add $10$ to both sides of the equation.

$7 x = - 3 + 10$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Add $- 3$ and $10$ .

$7 x = 7$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

$7 x = 7$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Divide each term in $7 x = 7$ by $7$ and simplify.

Divide each term in $7 x = 7$ by $7$ .

$\frac{7 x}{7} = \frac{7}{7}$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{7}{7}$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Divide $x$ by $1$ .

$x = \frac{7}{7}$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

$x = \frac{7}{7}$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

$x = \frac{7}{7}$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Simplify the right side.

Divide $7$ by $7$ .

$x = 1$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

$x = 1$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

$x = 1$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

$x = 1$

$- x + 3 y + 5 = 1$

$z = 5 - 3 x + 2 y$

Step 4

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

Replace all occurrences of $x$ in $- x + 3 y + 5 = 1$ with $1$ .

$- (1) + 3 y + 5 = 1$

$x = 1$

$z = 5 - 3 x + 2 y$

Simplify the left side.

Simplify $- (1) + 3 y + 5$ .

Multiply $- 1$ by $1$ .

$- 1 + 3 y + 5 = 1$

$x = 1$

$z = 5 - 3 x + 2 y$

Add $- 1$ and $5$ .

$3 y + 4 = 1$

$x = 1$

$z = 5 - 3 x + 2 y$

$3 y + 4 = 1$

$x = 1$

$z = 5 - 3 x + 2 y$

$3 y + 4 = 1$

$x = 1$

$z = 5 - 3 x + 2 y$

Replace all occurrences of $x$ in $z = 5 - 3 x + 2 y$ with $1$ .

$z = 5 - 3 \cdot 1 + 2 y$

$3 y + 4 = 1$

$x = 1$

Simplify the right side.

Simplify $5 - 3 \cdot 1 + 2 y$ .

Multiply $- 3$ by $1$ .

$z = 5 - 3 + 2 y$

$3 y + 4 = 1$

$x = 1$

Subtract $3$ from $5$ .

$z = 2 + 2 y$

$3 y + 4 = 1$

$x = 1$

$z = 2 + 2 y$

$3 y + 4 = 1$

$x = 1$

$z = 2 + 2 y$

$3 y + 4 = 1$

$x = 1$

$z = 2 + 2 y$

$3 y + 4 = 1$

$x = 1$

Step 5

Solve for $y$ in $3 y + 4 = 1$ .

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

Subtract $4$ from both sides of the equation.

$3 y = 1 - 4$

$z = 2 + 2 y$

$x = 1$

Subtract $4$ from $1$ .

$3 y = - 3$

$z = 2 + 2 y$

$x = 1$

$3 y = - 3$

$z = 2 + 2 y$

$x = 1$

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

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

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

$z = 2 + 2 y$

$x = 1$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

$z = 2 + 2 y$

$x = 1$

Divide $y$ by $1$ .

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

$z = 2 + 2 y$

$x = 1$

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

$z = 2 + 2 y$

$x = 1$

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

$z = 2 + 2 y$

$x = 1$

Simplify the right side.

Divide $- 3$ by $3$ .

$y = - 1$

$z = 2 + 2 y$

$x = 1$

$y = - 1$

$z = 2 + 2 y$

$x = 1$

$y = - 1$

$z = 2 + 2 y$

$x = 1$

$y = - 1$

$z = 2 + 2 y$

$x = 1$

Step 6

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

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

$z = 2 + 2 (- 1)$

$y = - 1$

$x = 1$

Simplify the right side.

Simplify $2 + 2 (- 1)$ .

Multiply $2$ by $- 1$ .

$z = 2 - 2$

$y = - 1$

$x = 1$

Subtract $2$ from $2$ .

$z = 0$

$y = - 1$

$x = 1$

$z = 0$

$y = - 1$

$x = 1$

$z = 0$

$y = - 1$

$x = 1$

$z = 0$

$y = - 1$

$x = 1$

Step 7

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

$(1, - 1, 0)$

Step 8

The result can be shown in multiple forms.

Point Form:

$(1, - 1, 0)$

Equation Form:

$x = 1, y = - 1, z = 0$

Solve the System of Equations 3x-2y+z=5 2x+y+z=1 x+4y-2z=-3

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