Solve {l}{2x+y=3}{x+y-z=12}{4x+3y+z=0}

$$\left\{ \begin{array}{l}{ 2 x + y = 3 }\\{ x + y - z = 12 }\\{ 4 x + 3 y + z = 0 }\end{array} \right.$$

Solve for x, y, z

$$x=0$$

$$y=3$$

$$z=-9$$

Short Steps Using Substitution

Solve 2x+y=3 for y.

$$y=-2x+3$$

Substitute -2x+3 for y in the second and third equation.

$$x-2x+3-z=12 4x+3\left(-2x+3\right)+z=0$$

Solve these equations for x and z respectively.

$$x=-9-z z=2x-9$$

Substitute -9-z for x in the equation z=2x-9.

$$z=2\left(-9-z\right)-9$$

Solve z=2\left(-9-z\right)-9 for z.

$$z=-9$$

Substitute -9 for z in the equation x=-9-z.

$$x=-9-\left(-9\right)$$

Calculate x from x=-9-\left(-9\right).

$$x=0$$

Substitute 0 for x in the equation y=-2x+3.

$$y=-2\times 0+3$$

Calculate y from y=-2\times 0+3.

$$y=3$$

The system is now solved.

$$x=0 y=3 z=-9$$