Solve the System of Equations y+z=4 x+z=8 x+2y+2z=12

$y + z = 4$ $x + z = 8$ $x + 2 y + 2 z = 12$

Subtract $z$ from both sides of the equation.

$y = 4 - z$

$x + z = 8$

$x + 2 y + 2 z = 12$

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

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

$x + 2 (4 - z) + 2 z = 12$

$y = 4 - z$

$x + z = 8$

Simplify the left side.

Simplify $x + 2 (4 - z) + 2 z$ .

Simplify each term.

Apply the distributive property.

$x + 2 \cdot 4 + 2 (- z) + 2 z = 12$

$y = 4 - z$

$x + z = 8$

Multiply $2$ by $4$ .

$x + 8 + 2 (- z) + 2 z = 12$

$y = 4 - z$

$x + z = 8$

Multiply $- 1$ by $2$ .

$x + 8 - 2 z + 2 z = 12$

$y = 4 - z$

$x + z = 8$

$x + 8 - 2 z + 2 z = 12$

$y = 4 - z$

$x + z = 8$

Combine the opposite terms in $x + 8 - 2 z + 2 z$ .

Add $- 2 z$ and $2 z$ .

$x + 8 + 0 = 12$

$y = 4 - z$

$x + z = 8$

Add $x + 8$ and $0$ .

$x + 8 = 12$

$y = 4 - z$

$x + z = 8$

$x + 8 = 12$

$y = 4 - z$

$x + z = 8$

$x + 8 = 12$

$y = 4 - z$

$x + z = 8$

$x + 8 = 12$

$y = 4 - z$

$x + z = 8$

$x + 8 = 12$

$y = 4 - z$

$x + z = 8$

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

Subtract $8$ from both sides of the equation.

$x = 12 - 8$

$y = 4 - z$

$x + z = 8$

Subtract $8$ from $12$ .

$x = 4$

$y = 4 - z$

$x + z = 8$

$x = 4$

$y = 4 - z$

$x + z = 8$

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

Replace all occurrences of $x$ in $x + z = 8$ with $4$ .

$(4) + z = 8$

$x = 4$

$y = 4 - z$

Simplify the left side.

Remove parentheses.

$4 + z = 8$

$x = 4$

$y = 4 - z$

$4 + z = 8$

$x = 4$

$y = 4 - z$

$4 + z = 8$

$x = 4$

$y = 4 - z$

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

Subtract $4$ from both sides of the equation.

$z = 8 - 4$

$x = 4$

$y = 4 - z$

Subtract $4$ from $8$ .

$z = 4$

$x = 4$

$y = 4 - z$

$z = 4$

$x = 4$

$y = 4 - z$

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

Replace all occurrences of $z$ in $y = 4 - z$ with $4$ .

$y = 4 - (4)$

$z = 4$

$x = 4$

Simplify the right side.

Simplify $4 - (4)$ .

Multiply $- 1$ by $4$ .

$y = 4 - 4$

$z = 4$

$x = 4$

Subtract $4$ from $4$ .

$y = 0$

$z = 4$

$x = 4$

$y = 0$

$z = 4$

$x = 4$

$y = 0$

$z = 4$

$x = 4$

$y = 0$

$z = 4$

$x = 4$

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

$(4, 0, 4)$

The result can be shown in multiple forms.

Point Form:

$(4, 0, 4)$

Equation Form:

$x = 4, y = 0, z = 4$

Solve the System of Equations y+z=4 x+z=8 x+2y+2z=12

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