Solve {l}{x+y-40=0}{z=x-2y+14}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x + y - 40 = 0 }\\ { z = x - 2 y + 14 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{a}{3}+22$$

$$y=-\frac{a}{3}+18$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=x-2y+14 x+y-40=0 a=z$$

Substitute x-2y+14 for z in the equation a=z.

$$a=x-2y+14$$

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

$$y=-x+40 x=2y+a-14$$

Substitute -x+40 for y in the equation x=2y+a-14.

$$x=2\left(-x+40\right)+a-14$$

Solve x=2\left(-x+40\right)+a-14 for x.

$$x=22+\frac{1}{3}a$$

Substitute 22+\frac{1}{3}a for x in the equation y=-x+40.

$$y=-\left(22+\frac{1}{3}a\right)+40$$

Calculate y from y=-\left(22+\frac{1}{3}a\right)+40.

$$y=18-\frac{1}{3}a$$

Substitute 18-\frac{1}{3}a for y and 22+\frac{1}{3}a for x in the equation z=x-2y+14.

$$z=22+\frac{1}{3}a-2\left(18-\frac{1}{3}a\right)+14$$

Calculate z from z=22+\frac{1}{3}a-2\left(18-\frac{1}{3}a\right)+14.

$$z=a$$

The system is now solved.

$$x=22+\frac{1}{3}a y=18-\frac{1}{3}a z=a$$