Solve {l}{5x+3y+z=1}{3x+2y+z=4}{6x+5y+3z=6}

$$\left. \begin{array} { l } { 5 x + 3 y + z = 1 } \\ { 3 x + 2 y + z = 4 } \\ { 6 x + 5 y + 3 z = 6 } \end{array} \right.$$

Solve for x, y, z

$$x=9$$

$$y=-21$$

$$z=19$$

Short Steps Using Substitution

Solve 5x+3y+z=1 for z.

$$z=-5x-3y+1$$

Substitute -5x-3y+1 for z in the second and third equation.

$$3x+2y-5x-3y+1=4 6x+5y+3\left(-5x-3y+1\right)=6$$

Solve these equations for y and x respectively.

$$y=-2x-3 x=-\frac{4}{9}y-\frac{1}{3}$$

Substitute -2x-3 for y in the equation x=-\frac{4}{9}y-\frac{1}{3}.

$$x=-\frac{4}{9}\left(-2x-3\right)-\frac{1}{3}$$

Solve x=-\frac{4}{9}\left(-2x-3\right)-\frac{1}{3} for x.

$$x=9$$

Substitute 9 for x in the equation y=-2x-3.

$$y=-2\times 9-3$$

Calculate y from y=-2\times 9-3.

$$y=-21$$

Substitute -21 for y and 9 for x in the equation z=-5x-3y+1.

$$z=-5\times 9-3\left(-21\right)+1$$

Calculate z from z=-5\times 9-3\left(-21\right)+1.

$$z=19$$

The system is now solved.

$$x=9 y=-21 z=19$$