Solve {c}{x+y+z=1}{2x+4y+2z=10}{-x+8y-3z=33}

$$\left. \begin{array} { c } { x + y + z = 1 } \\ { 2 x + 4 y + 2 z = 10 } \\ { - x + 8 y - 3 z = 33 } \end{array} \right.$$

Solve for x, y, z

$$x=-4$$

$$y=4$$

$$z=1$$

Short Steps Using Substitution

Solve x+y+z=1 for x.

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

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

$$2\left(-y-z+1\right)+4y+2z=10 -\left(-y-z+1\right)+8y-3z=33$$

Solve these equations for y and z respectively.

$$y=4 z=\frac{9}{2}y-17$$

Substitute 4 for y in the equation z=\frac{9}{2}y-17.

$$z=\frac{9}{2}\times 4-17$$

Calculate z from z=\frac{9}{2}\times 4-17.

$$z=1$$

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

$$x=-4-1+1$$

Calculate x from x=-4-1+1.

$$x=-4$$

The system is now solved.

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