Solve {c}{4x+y-z=1}{x+4y+4z=1}{x+y+z=1}

$$\left\{ \begin{array} { c } { 4 x + y - z = 1 } \\ { x + 4 y + 4 z = 1 } \\ { x + y + z = 1 } \end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

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

Short Steps Using Substitution

Solve 4x+y-z=1 for y.

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

Substitute -4x+z+1 for y in the second and third equation.

$$x+4\left(-4x+z+1\right)+4z=1 x-4x+z+1+z=1$$

Solve these equations for x and z respectively.

$$x=\frac{8}{15}z+\frac{1}{5} z=\frac{3}{2}x$$

Substitute \frac{8}{15}z+\frac{1}{5} for x in the equation z=\frac{3}{2}x.

$$z=\frac{3}{2}\left(\frac{8}{15}z+\frac{1}{5}\right)$$

Solve z=\frac{3}{2}\left(\frac{8}{15}z+\frac{1}{5}\right) for z.

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

Substitute \frac{3}{2} for z in the equation x=\frac{8}{15}z+\frac{1}{5}.

$$x=\frac{8}{15}\times \frac{3}{2}+\frac{1}{5}$$

Calculate x from x=\frac{8}{15}\times \frac{3}{2}+\frac{1}{5}.

$$x=1$$

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

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

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

$$y=-\frac{3}{2}$$

The system is now solved.

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