Solve {l}{x+3y=6500}{15y=6779}{z=3}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x + 3 y = 6500 }\\ { 15 y = 6779 }\\ { z = 3 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z, a

$$a=3$$

Solution Steps

Consider the second equation. Divide both sides by 15.

$$y=\frac{6779}{15}$$

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

$$x+3\times \frac{6779}{15}=6500$$

Multiply 3 and \frac{6779}{15} to get \frac{6779}{5}.

$$x+\frac{6779}{5}=6500$$

Subtract \frac{6779}{5} from both sides.

$$x=6500-\frac{6779}{5}$$

Subtract \frac{6779}{5} from 6500 to get \frac{25721}{5}.

$$x=\frac{25721}{5}$$

The system is now solved.

$$x=\frac{25721}{5} y=\frac{6779}{15} z=3 a=3$$