Solve {l}{x-y=-1}{z=x+2y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x - y = -1 }\\ { z = x + 2 y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{a-2}{3}$$

$$y=\frac{a+1}{3}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=x+2y x-y=-1 a=z$$

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

$$a=x+2y$$

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

$$y=x+1 x=a-2y$$

Substitute x+1 for y in the equation x=a-2y.

$$x=a-2\left(x+1\right)$$

Solve x=a-2\left(x+1\right) for x.

$$x=-\frac{2}{3}+\frac{1}{3}a$$

Substitute -\frac{2}{3}+\frac{1}{3}a for x in the equation y=x+1.

$$y=-\frac{2}{3}+\frac{1}{3}a+1$$

Calculate y from y=-\frac{2}{3}+\frac{1}{3}a+1.

$$y=\frac{1}{3}+\frac{1}{3}a$$

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

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

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

$$z=a$$

The system is now solved.

$$x=-\frac{2}{3}+\frac{1}{3}a y=\frac{1}{3}+\frac{1}{3}a z=a$$