Solve {l}{8x+4y+z=1}{92+3y=8}{3x+yy+z=0}

$$\left\{ \begin{array} { l } { 8 x + 4 y + z = 1 } \\ { 92 + 3 y = 8 } \\ { 3 x + y y + z = 0 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{897}{5} = 179\frac{2}{5} = 179.4$$

$$y=-28$$

$$z = -\frac{6611}{5} = -1322\frac{1}{5} = -1322.2$$

Short Steps Using Substitution

Solve 8x+4y+z=1 for z.

$$z=-8x-4y+1$$

Substitute -8x-4y+1 for z in the equation 3x+yy+z=0.

$$3x+yy-8x-4y+1=0$$

Solve the second equation for y and the third equation for x.

$$y=-28 x=\frac{1}{5}y^{2}-\frac{4}{5}y+\frac{1}{5}$$

Substitute -28 for y in the equation x=\frac{1}{5}y^{2}-\frac{4}{5}y+\frac{1}{5}.

$$x=\frac{1}{5}\left(-28\right)^{2}-\frac{4}{5}\left(-28\right)+\frac{1}{5}$$

Calculate x from x=\frac{1}{5}\left(-28\right)^{2}-\frac{4}{5}\left(-28\right)+\frac{1}{5}.

$$x=\frac{897}{5}$$

Substitute -28 for y and \frac{897}{5} for x in the equation z=-8x-4y+1.

$$z=-8\times \frac{897}{5}-4\left(-28\right)+1$$

Calculate z from z=-8\times \frac{897}{5}-4\left(-28\right)+1.

$$z=-\frac{6611}{5}$$

The system is now solved.

$$x=\frac{897}{5} y=-28 z=-\frac{6611}{5}$$