Solve {l}{x+2y+3z=4}{y-z+x=3}{3x+y+z=7}

$$\left. \begin{array} { l } { x + 2 y + 3 z = 4 } \\ { y - z + x = 3 } \\ { 3 x + y + z = 7 } \end{array} \right.$$

Solve for x, y, z

$$x=2$$

$$y=1$$

$$z=0$$

Short Steps Using Substitution

Solve x+2y+3z=4 for x.

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

Substitute -2y-3z+4 for x in the second and third equation.

$$y-z-2y-3z+4=3 3\left(-2y-3z+4\right)+y+z=7$$

Solve these equations for y and z respectively.

$$y=-4z+1 z=\frac{5}{8}-\frac{5}{8}y$$

Substitute -4z+1 for y in the equation z=\frac{5}{8}-\frac{5}{8}y.

$$z=\frac{5}{8}-\frac{5}{8}\left(-4z+1\right)$$

Solve z=\frac{5}{8}-\frac{5}{8}\left(-4z+1\right) for z.

$$z=0$$

Substitute 0 for z in the equation y=-4z+1.

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

Calculate y from y=-4\times 0+1.

$$y=1$$

Substitute 1 for y and 0 for z in the equation x=-2y-3z+4.

$$x=-2-3\times 0+4$$

Calculate x from x=-2-3\times 0+4.

$$x=2$$

The system is now solved.

$$x=2 y=1 z=0$$