Solve {l}{40+0.3x=50t}{y=0.25x}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 40 + 0.3 x = 50 t }\\ { y = 0.25 x }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=\frac{125t-100}{3}$$

Short Steps Using Substitution

Reorder the equations.

$$y=0.25x 40+0.3x=50t z=y$$

Substitute 0.25x for y in the equation z=y.

$$z=0.25x$$

Solve 40+0.3x=50t for x.

$$x=-\frac{400}{3}+\frac{500}{3}t$$

Substitute -\frac{400}{3}+\frac{500}{3}t for x in the equation z=0.25x.

$$z=0.25\left(-\frac{400}{3}+\frac{500}{3}t\right)$$

Calculate z from z=0.25\left(-\frac{400}{3}+\frac{500}{3}t\right).

$$z=-\frac{100}{3}+\frac{125}{3}t$$

Substitute -\frac{400}{3}+\frac{500}{3}t for x in the equation y=0.25x.

$$y=0.25\left(-\frac{400}{3}+\frac{500}{3}t\right)$$

Calculate y from y=0.25\left(-\frac{400}{3}+\frac{500}{3}t\right).

$$y=-\frac{100}{3}+\frac{125}{3}t$$

The system is now solved.

$$x=-\frac{400}{3}+\frac{500}{3}t y=-\frac{100}{3}+\frac{125}{3}t z=-\frac{100}{3}+\frac{125}{3}t$$