Solve {l}{x=1}{y=2000x+{(10-x)}1800}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { x = 1 }\\ { y = 2000 x + {(10 - x)} 1800 }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=18200$$

Solution Steps

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

$$y=2000\times 1+\left(10-1\right)\times 1800$$

Multiply 2000 and 1 to get 2000.

$$y=2000+\left(10-1\right)\times 1800$$

Subtract 1 from 10 to get 9.

$$y=2000+9\times 1800$$

Multiply 9 and 1800 to get 16200.

$$y=2000+16200$$

Add 2000 and 16200 to get 18200.

$$y=18200$$

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

$$z=18200$$

The system is now solved.

$$x=1 y=18200 z=18200$$