Solve {l}{x=32,0}{x+y+2z=8}{2x+3y+z=16}

$$\left\{ \begin{array} { l } { x = 32,0 } \\ { x + y + 2 z = 8 } \\ { 2 x + 3 y + z = 16 } \end{array} \right.$$

Solve for x, y, z

$$x=32$$

$$y=-\frac{72}{5}=-14,4$$

$$z=-\frac{24}{5}=-4,8$$

Short Steps Using Substitution

Substitute 32 for x in the second and third equation.

$$32+y+2z=8 2\times 32+3y+z=16$$

Solve these equations for y and z respectively.

$$y=-24-2z z=-48-3y$$

Substitute -24-2z for y in the equation z=-48-3y.

$$z=-48-3\left(-24-2z\right)$$

Solve z=-48-3\left(-24-2z\right) for z.

$$z=-\frac{24}{5}$$

Substitute -\frac{24}{5} for z in the equation y=-24-2z.

$$y=-24-2\left(-\frac{24}{5}\right)$$

Calculate y from y=-24-2\left(-\frac{24}{5}\right).

$$y=-\frac{72}{5}$$

The system is now solved.

$$x=32 y=-\frac{72}{5} z=-\frac{24}{5}$$