Solve (1+x)(1+y)(1+z)=2010

$$(1+x)(1+y)(1+z)=2010$$

Solve for x

$$x=-\frac{yz+y+z-2009}{yz+y+z+1}$$

$$z\neq -1\text{ and }y\neq -1$$

Steps for Solving Linear Equation

Use the distributive property to multiply 1+x by 1+y.

$$\left(1+y+x+xy\right)\left(1+z\right)=2010$$

Use the distributive property to multiply 1+y+x+xy by 1+z.

$$1+z+y+yz+x+xz+xy+xyz=2010$$

Subtract 1 from both sides.

$$z+y+yz+x+xz+xy+xyz=2010-1$$

Subtract 1 from 2010 to get 2009.

$$z+y+yz+x+xz+xy+xyz=2009$$

Subtract z from both sides.

$$y+yz+x+xz+xy+xyz=2009-z$$

Subtract y from both sides.

$$yz+x+xz+xy+xyz=2009-z-y$$

Subtract yz from both sides.

$$x+xz+xy+xyz=2009-z-y-yz$$

Combine all terms containing x.

$$\left(1+z+y+yz\right)x=2009-z-y-yz$$

The equation is in standard form.

$$\left(yz+y+z+1\right)x=2009-z-y-yz$$

Divide both sides by yz+y+z+1.

$$\frac{\left(yz+y+z+1\right)x}{yz+y+z+1}=\frac{2009-z-y-yz}{yz+y+z+1}$$

Dividing by yz+y+z+1 undoes the multiplication by yz+y+z+1.

$$x=\frac{2009-z-y-yz}{yz+y+z+1}$$

Solve for y

$$y=-\frac{xz+x+z-2009}{xz+x+z+1}$$

$$z\neq -1\text{ and }x\neq -1$$

Steps for Solving Linear Equation

Use the distributive property to multiply 1+x by 1+y.

$$\left(1+y+x+xy\right)\left(1+z\right)=2010$$

Use the distributive property to multiply 1+y+x+xy by 1+z.

$$1+z+y+yz+x+xz+xy+xyz=2010$$

Subtract 1 from both sides.

$$z+y+yz+x+xz+xy+xyz=2010-1$$

Subtract 1 from 2010 to get 2009.

$$z+y+yz+x+xz+xy+xyz=2009$$

Subtract z from both sides.

$$y+yz+x+xz+xy+xyz=2009-z$$

Subtract x from both sides.

$$y+yz+xz+xy+xyz=2009-z-x$$

Subtract xz from both sides.

$$y+yz+xy+xyz=2009-z-x-xz$$

Combine all terms containing y.

$$\left(1+z+x+xz\right)y=2009-z-x-xz$$

The equation is in standard form.

$$\left(xz+x+z+1\right)y=2009-z-x-xz$$

Divide both sides by xz+x+z+1.

$$\frac{\left(xz+x+z+1\right)y}{xz+x+z+1}=\frac{2009-z-x-xz}{xz+x+z+1}$$

Dividing by xz+x+z+1 undoes the multiplication by xz+x+z+1.

$$y=\frac{2009-z-x-xz}{xz+x+z+1}$$