Solve {l}{x=6y}{x-4=10y-20z}{x+12=3y+18z}

$$\left\{ \begin{array} { l } { x = 6 y } \\ { x - 4 = 10 y - 20 z } \\ { x + 12 = 3 y + 18 z } \end{array} \right.$$

Solve for x, y, z

$$x=84$$

$$y=14$$

$$z=3$$

Short Steps Using Substitution

Substitute 6y for x in the second and third equation.

$$6y-4=10y-20z 6y+12=3y+18z$$

Solve these equations for y and z respectively.

$$y=-1+5z z=\frac{2}{3}+\frac{1}{6}y$$

Substitute -1+5z for y in the equation z=\frac{2}{3}+\frac{1}{6}y.

$$z=\frac{2}{3}+\frac{1}{6}\left(-1+5z\right)$$

Solve z=\frac{2}{3}+\frac{1}{6}\left(-1+5z\right) for z.

$$z=3$$

Substitute 3 for z in the equation y=-1+5z.

$$y=-1+5\times 3$$

Calculate y from y=-1+5\times 3.

$$y=14$$

Substitute 14 for y in the equation x=6y.

$$x=6\times 14$$

Calculate x from x=6\times 14.

$$x=84$$

The system is now solved.

$$x=84 y=14 z=3$$