Solve {l}{x=5/2}{y={(10+x)}{(150-10x)}}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { x = \frac{5}{2} }\\ { y = {(10 + x)} {(150 - 10 x)} }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z = \frac{3125}{2} = 1562\frac{1}{2} = 1562.5$$

Solution Steps

Consider the second equation. Insert the known values of variables into the equation.

$$y=\left(10+\frac{5}{2}\right)\left(150-10\times \frac{5}{2}\right)$$

Add 10 and \frac{5}{2} to get \frac{25}{2}.

$$y=\frac{25}{2}\left(150-10\times \frac{5}{2}\right)$$

Multiply -10 and \frac{5}{2} to get -25.

$$y=\frac{25}{2}\left(150-25\right)$$

Subtract 25 from 150 to get 125.

$$y=\frac{25}{2}\times 125$$

Multiply \frac{25}{2} and 125 to get \frac{3125}{2}.

$$y=\frac{3125}{2}$$

Consider the third equation. Insert the known values of variables into the equation.

$$z=\frac{3125}{2}$$

The system is now solved.

$$x=\frac{5}{2} y=\frac{3125}{2} z=\frac{3125}{2}$$