Solve {r}{4x+y-z=0}{x+z-z=0}{100-x+y=0}

$$\left. \begin{array} { r } { 4 x + y - z = 0 } \\ { x + z - z = 0 } \\ { 100 - x + y = 0 } \end{array} \right.$$

Solve for x, y, z

$$x=0$$

$$y=-100$$

$$z=-100$$

Solution Steps

Consider the second equation. Combine z and -z to get 0.

$$x=0$$

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

$$100-0+y=0$$

Multiply -1 and 0 to get 0.

$$100+0+y=0$$

Add 100 and 0 to get 100.

$$100+y=0$$

Subtract 100 from both sides. Anything subtracted from zero gives its negation.

$$y=-100$$

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

$$4\times 0-100-z=0$$

Multiply 4 and 0 to get 0.

$$0-100-z=0$$

Subtract 100 from 0 to get -100.

$$-100-z=0$$

Add 100 to both sides. Anything plus zero gives itself.

$$-z=100$$

Divide both sides by -1.

$$z=-100$$

The system is now solved.

$$x=0 y=-100 z=-100$$