Solve {l}{x=2y-3}{2y=3z}{x+y+z=41}

$$\left. \begin{array} { l } { x = 2 y - 3 } \\ { 2 y = 3 z } \\ { x + y + z = 41 } \end{array} \right.$$

Solve for x, y, z

$$x=21$$

$$y=12$$

$$z=8$$

Short Steps Using Substitution

Substitute 2y-3 for x in the equation x+y+z=41.

$$2y-3+y+z=41$$

Solve the second equation for y and the third equation for z.

$$y=\frac{3}{2}z z=-3y+44$$

Substitute \frac{3}{2}z for y in the equation z=-3y+44.

$$z=-3\times \frac{3}{2}z+44$$

Solve z=-3\times \frac{3}{2}z+44 for z.

$$z=8$$

Substitute 8 for z in the equation y=\frac{3}{2}z.

$$y=\frac{3}{2}\times 8$$

Calculate y from y=\frac{3}{2}\times 8.

$$y=12$$

Substitute 12 for y in the equation x=2y-3.

$$x=2\times 12-3$$

Calculate x from x=2\times 12-3.

$$x=21$$

The system is now solved.

$$x=21 y=12 z=8$$