Solve {l}{2x+12y+20z=272}{x+y+z=52}{2x+12y+z=158}

$$\left. \begin{array} { l } { 2 x + 12 y + 20 z = 272 } \\ { x + y + z = 52 } \\ { 2 x + 12 y + z = 158 } \end{array} \right.$$

Solve for x, y, z

$$x=40$$

$$y=6$$

$$z=6$$

Short Steps Using Substitution

Reorder the equations.

$$x+y+z=52 2x+12y+20z=272 2x+12y+z=158$$

Solve x+y+z=52 for x.

$$x=-y-z+52$$

Substitute -y-z+52 for x in the second and third equation.

$$2\left(-y-z+52\right)+12y+20z=272 2\left(-y-z+52\right)+12y+z=158$$

Solve these equations for y and z respectively.

$$y=\frac{84}{5}-\frac{9}{5}z z=10y-54$$

Substitute \frac{84}{5}-\frac{9}{5}z for y in the equation z=10y-54.

$$z=10\left(\frac{84}{5}-\frac{9}{5}z\right)-54$$

Solve z=10\left(\frac{84}{5}-\frac{9}{5}z\right)-54 for z.

$$z=6$$

Substitute 6 for z in the equation y=\frac{84}{5}-\frac{9}{5}z.

$$y=\frac{84}{5}-\frac{9}{5}\times 6$$

Calculate y from y=\frac{84}{5}-\frac{9}{5}\times 6.

$$y=6$$

Substitute 6 for y and 6 for z in the equation x=-y-z+52.

$$x=-6-6+52$$

Calculate x from x=-6-6+52.

$$x=40$$

The system is now solved.

$$x=40 y=6 z=6$$