Solve {l}{x^{2}+y^{2}+(z-1)^{2}=x^{2}+(y-2)^{2}+z^{2}}{(x-2)^{2}+y^{2}+z^{2}=x^{2}+(y-2)^{2}+z^{2}}{x^{2}+y^{2}+z^2=x^2+y^2+(z-1)^2}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } + ( z - 1 ) ^ { 2 } = x ^ { 2 } + ( y - 2 ) ^ { 2 } + z ^ { 2 } } \\ { ( x - 2 ) ^ { 2 } + y ^ { 2 } + z ^ { 2 } = x ^ { 2 } + ( y - 2 ) ^ { 2 } + z ^ { 2 } } \\ { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = x ^ { 2 } + y ^ { 2 } + ( z - 1 ) ^ { 2 } } \end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y=1$$

$$z=\frac{1}{2}=0.5$$

Solution Steps

Consider the first equation. Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(z-1\right)^{2}.

$$x^{2}+y^{2}+z^{2}-2z+1=x^{2}+\left(y-2\right)^{2}+z^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-2\right)^{2}.

$$x^{2}+y^{2}+z^{2}-2z+1=x^{2}+y^{2}-4y+4+z^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+y^{2}+z^{2}-2z+1-x^{2}=y^{2}-4y+4+z^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$y^{2}+z^{2}-2z+1=y^{2}-4y+4+z^{2}$$

Subtract y^{2} from both sides.

$$y^{2}+z^{2}-2z+1-y^{2}=-4y+4+z^{2}$$

Combine y^{2} and -y^{2} to get 0.

$$z^{2}-2z+1=-4y+4+z^{2}$$

Add 4y to both sides.

$$z^{2}-2z+1+4y=4+z^{2}$$

Subtract z^{2} from both sides.

$$z^{2}-2z+1+4y-z^{2}=4$$

Combine z^{2} and -z^{2} to get 0.

$$-2z+1+4y=4$$

Subtract 1 from both sides.

$$-2z+4y=4-1$$

Subtract 1 from 4 to get 3.

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

Consider the second equation. Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$x^{2}-4x+4+y^{2}+z^{2}=x^{2}+\left(y-2\right)^{2}+z^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-2\right)^{2}.

$$x^{2}-4x+4+y^{2}+z^{2}=x^{2}+y^{2}-4y+4+z^{2}$$

Subtract x^{2} from both sides.

$$x^{2}-4x+4+y^{2}+z^{2}-x^{2}=y^{2}-4y+4+z^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$-4x+4+y^{2}+z^{2}=y^{2}-4y+4+z^{2}$$

Subtract y^{2} from both sides.

$$-4x+4+y^{2}+z^{2}-y^{2}=-4y+4+z^{2}$$

Combine y^{2} and -y^{2} to get 0.

$$-4x+4+z^{2}=-4y+4+z^{2}$$

Add 4y to both sides.

$$-4x+4+z^{2}+4y=4+z^{2}$$

Subtract z^{2} from both sides.

$$-4x+4+z^{2}+4y-z^{2}=4$$

Combine z^{2} and -z^{2} to get 0.

$$-4x+4+4y=4$$

Subtract 4 from both sides.

$$-4x+4y=4-4$$

Subtract 4 from 4 to get 0.

$$-4x+4y=0$$

Consider the third equation. Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(z-1\right)^{2}.

$$x^{2}+y^{2}+z^{2}=x^{2}+y^{2}+z^{2}-2z+1$$

Subtract x^{2} from both sides.

$$x^{2}+y^{2}+z^{2}-x^{2}=y^{2}+z^{2}-2z+1$$

Combine x^{2} and -x^{2} to get 0.

$$y^{2}+z^{2}=y^{2}+z^{2}-2z+1$$

Subtract y^{2} from both sides.

$$y^{2}+z^{2}-y^{2}=z^{2}-2z+1$$

Combine y^{2} and -y^{2} to get 0.

$$z^{2}=z^{2}-2z+1$$

Subtract z^{2} from both sides.

$$z^{2}-z^{2}=-2z+1$$

Combine z^{2} and -z^{2} to get 0.

$$0=-2z+1$$

Swap sides so that all variable terms are on the left hand side.

$$-2z+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-2z=-1$$

Divide both sides by -2.

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

Fraction \frac{-1}{-2} can be simplified to \frac{1}{2} by removing the negative sign from both the numerator and the denominator.

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

Consider the first equation. Insert the known values of variables into the equation.

$$-2\times \frac{1}{2}+4y=3$$

Multiply -2 and \frac{1}{2} to get -1.

$$-1+4y=3$$

Add 1 to both sides.

$$4y=3+1$$

Add 3 and 1 to get 4.

$$4y=4$$

Divide both sides by 4.

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

Divide 4 by 4 to get 1.

$$y=1$$

Consider the second equation. Insert the known values of variables into the equation.

$$-4x+4\times 1=0$$

Multiply 4 and 1 to get 4.

$$-4x+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-4x=-4$$

Divide both sides by -4.

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

Divide -4 by -4 to get 1.

$$x=1$$

The system is now solved.

$$x=1 y=1 z=\frac{1}{2}$$