Solve {l}{x+y=6}{z=4x-y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x + y = 6 }\\ { z = 4 x - y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{a+6}{5}$$

$$y=\frac{24-a}{5}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=4x-y x+y=6 a=z$$

Substitute 4x-y for z in the equation a=z.

$$a=4x-y$$

Solve the second equation for y and the third equation for x.

$$y=-x+6 x=\frac{1}{4}a+\frac{1}{4}y$$

Substitute -x+6 for y in the equation x=\frac{1}{4}a+\frac{1}{4}y.

$$x=\frac{1}{4}a+\frac{1}{4}\left(-x+6\right)$$

Solve x=\frac{1}{4}a+\frac{1}{4}\left(-x+6\right) for x.

$$x=\frac{6}{5}+\frac{1}{5}a$$

Substitute \frac{6}{5}+\frac{1}{5}a for x in the equation y=-x+6.

$$y=-\left(\frac{6}{5}+\frac{1}{5}a\right)+6$$

Calculate y from y=-\left(\frac{6}{5}+\frac{1}{5}a\right)+6.

$$y=\frac{24}{5}-\frac{1}{5}a$$

Substitute \frac{24}{5}-\frac{1}{5}a for y and \frac{6}{5}+\frac{1}{5}a for x in the equation z=4x-y.

$$z=4\left(\frac{6}{5}+\frac{1}{5}a\right)-\left(\frac{24}{5}-\frac{1}{5}a\right)$$

Calculate z from z=4\left(\frac{6}{5}+\frac{1}{5}a\right)-\left(\frac{24}{5}-\frac{1}{5}a\right).

$$z=a$$

The system is now solved.

$$x=\frac{6}{5}+\frac{1}{5}a y=\frac{24}{5}-\frac{1}{5}a z=a$$