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

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

Solve for x, y, z

$$x=1$$

$$y=-5$$

$$z=5$$

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.

$$-y-z+1+2y+3z=6 -y-z+1+3y+4z=6$$

Solve these equations for y and z respectively.

$$y=-2z+5 z=-\frac{2}{3}y+\frac{5}{3}$$

Substitute -2z+5 for y in the equation z=-\frac{2}{3}y+\frac{5}{3}.

$$z=-\frac{2}{3}\left(-2z+5\right)+\frac{5}{3}$$

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

$$z=5$$

Substitute 5 for z in the equation y=-2z+5.

$$y=-2\times 5+5$$

Calculate y from y=-2\times 5+5.

$$y=-5$$

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

$$x=-\left(-5\right)-5+1$$

Calculate x from x=-\left(-5\right)-5+1.

$$x=1$$

The system is now solved.

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