Solve {l}{3x+2y+z=196}{x+3y+2z=200}{2x+y+3z=168}

$$\left\{ \begin{array} { l } { 3 x + 2 y + z = 196 } \\ { x + 3 y + 2 z = 200 } \\ { 2 x + y + 3 z = 168 } \end{array} \right.$$

Solve for x, y, z

$$x=30$$

$$y=42$$

$$z=22$$

Short Steps Using Substitution

Solve 3x+2y+z=196 for z.

$$z=-3x-2y+196$$

Substitute -3x-2y+196 for z in the second and third equation.

$$x+3y+2\left(-3x-2y+196\right)=200 2x+y+3\left(-3x-2y+196\right)=168$$

Solve these equations for y and x respectively.

$$y=-5x+192 x=60-\frac{5}{7}y$$

Substitute -5x+192 for y in the equation x=60-\frac{5}{7}y.

$$x=60-\frac{5}{7}\left(-5x+192\right)$$

Solve x=60-\frac{5}{7}\left(-5x+192\right) for x.

$$x=30$$

Substitute 30 for x in the equation y=-5x+192.

$$y=-5\times 30+192$$

Calculate y from y=-5\times 30+192.

$$y=42$$

Substitute 42 for y and 30 for x in the equation z=-3x-2y+196.

$$z=-3\times 30-2\times 42+196$$

Calculate z from z=-3\times 30-2\times 42+196.

$$z=22$$

The system is now solved.

$$x=30 y=42 z=22$$