Solve {l}{x+y=84}{4y=1242}{z=18x+14}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x + y = 84 }\\ { 4 y = 1242 }\\ { z = 18 x + 14 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z, a

$$a=-4063$$

Solution Steps

Consider the second equation. Divide both sides by 4.

$$y=\frac{1242}{4}$$

Reduce the fraction \frac{1242}{4} to lowest terms by extracting and canceling out 2.

$$y=\frac{621}{2}$$

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

$$x+\frac{621}{2}=84$$

Subtract \frac{621}{2} from both sides.

$$x=84-\frac{621}{2}$$

Subtract \frac{621}{2} from 84 to get -\frac{453}{2}.

$$x=-\frac{453}{2}$$

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

$$z=18\left(-\frac{453}{2}\right)+14$$

Multiply 18 and -\frac{453}{2} to get -4077.

$$z=-4077+14$$

Add -4077 and 14 to get -4063.

$$z=-4063$$

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

$$a=-4063$$

The system is now solved.

$$x=-\frac{453}{2} y=\frac{621}{2} z=-4063 a=-4063$$