Solve {l}{text{(I)}x-y+z=4}{text{(II)}3x+y+z=1}{text{(III)}9x-3y-z=9}

$$\left. \begin{array} { l } { \text { (I) } x - y + z = 4 } \\ { \text { (II) } 3 x + y + z = 1 } \\ { \text { (III) } 9 x - 3 y - z = 9 } \end{array} \right.$$

Solve for x, y, z

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

$$y=-2$$

$$z = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Short Steps Using Substitution

Solve x-y+z=4 for x.

$$x=y-z+4$$

Substitute y-z+4 for x in the second and third equation.

$$3\left(y-z+4\right)+y+z=1 9\left(y-z+4\right)-3y-z=9$$

Solve these equations for y and z respectively.

$$y=\frac{1}{2}z-\frac{11}{4} z=\frac{27}{10}+\frac{3}{5}y$$

Substitute \frac{1}{2}z-\frac{11}{4} for y in the equation z=\frac{27}{10}+\frac{3}{5}y.

$$z=\frac{27}{10}+\frac{3}{5}\left(\frac{1}{2}z-\frac{11}{4}\right)$$

Solve z=\frac{27}{10}+\frac{3}{5}\left(\frac{1}{2}z-\frac{11}{4}\right) for z.

$$z=\frac{3}{2}$$

Substitute \frac{3}{2} for z in the equation y=\frac{1}{2}z-\frac{11}{4}.

$$y=\frac{1}{2}\times \frac{3}{2}-\frac{11}{4}$$

Calculate y from y=\frac{1}{2}\times \frac{3}{2}-\frac{11}{4}.

$$y=-2$$

Substitute -2 for y and \frac{3}{2} for z in the equation x=y-z+4.

$$x=-2-\frac{3}{2}+4$$

Calculate x from x=-2-\frac{3}{2}+4.

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

The system is now solved.

$$x=\frac{1}{2} y=-2 z=\frac{3}{2}$$