Solve {l}{y+z=6}{2x+y-4z=m}{2y-z=m}

$$\left\{ \begin{array} { l } { y + z = 6 } \\ { 2 x + y - 4 z = m } \\ { 2 y - z = m } \end{array} \right.$$

Solve for x, y, z

$$x=-\frac{m}{3}+7$$

$$y=\frac{m}{3}+2$$

$$z=-\frac{m}{3}+4$$

Short Steps Using Substitution

Solve y+z=6 for y.

$$y=-z+6$$

Substitute -z+6 for y in the second and third equation.

$$2x-z+6-4z=m 2\left(-z+6\right)-z=m$$

Solve these equations for x and z respectively.

$$x=\frac{5}{2}z-3+\frac{1}{2}m z=4-\frac{1}{3}m$$

Substitute 4-\frac{1}{3}m for z in the equation x=\frac{5}{2}z-3+\frac{1}{2}m.

$$x=\frac{5}{2}\left(4-\frac{1}{3}m\right)-3+\frac{1}{2}m$$

Calculate x from x=\frac{5}{2}\left(4-\frac{1}{3}m\right)-3+\frac{1}{2}m.

$$x=7-\frac{1}{3}m$$

Substitute 4-\frac{1}{3}m for z in the equation y=-z+6.

$$y=-\left(4-\frac{1}{3}m\right)+6$$

Calculate y from y=-\left(4-\frac{1}{3}m\right)+6.

$$y=2+\frac{1}{3}m$$

The system is now solved.

$$x=7-\frac{1}{3}m y=2+\frac{1}{3}m z=4-\frac{1}{3}m$$