Solve 2x+4y+3z+3z=x+2y+3z+0

$$2 x + 4 y + 3 z + 3 z = x + 2 y + 3 z + 0$$

Solve for x

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

Solution Steps

Combine 3z and 3z to get 6z.

$$2x+4y+6z=x+2y+3z+0$$

Anything plus zero gives itself.

$$2x+4y+6z=x+2y+3z$$

Subtract x from both sides.

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

Combine 2x and -x to get x.

$$x+4y+6z=2y+3z$$

Subtract 4y from both sides.

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

Combine 2y and -4y to get -2y.

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

Subtract 6z from both sides.

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

Combine 3z and -6z to get -3z.

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

Solve for y

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

Steps for Solving Linear Equation

Combine 3z and 3z to get 6z.

$$2x+4y+6z=x+2y+3z+0$$

Anything plus zero gives itself.

$$2x+4y+6z=x+2y+3z$$

Subtract 2y from both sides.

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

Combine 4y and -2y to get 2y.

$$2x+2y+6z=x+3z$$

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Subtract 6z from both sides.

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

Combine 3z and -6z to get -3z.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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