Solve {l}{2x-y-2=0}{z=x+1+y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 2 x - y - 2 = 0 }\\ { z = x + 1 + y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

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

$$y=\frac{2\left(a-2\right)}{3}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=x+1+y 2x-y-2=0 a=z$$

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

$$a=x+1+y$$

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

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

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

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

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

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

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

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

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

$$y=-\frac{4}{3}+\frac{2}{3}a$$

Substitute -\frac{4}{3}+\frac{2}{3}a for y and \frac{1}{3}+\frac{1}{3}a for x in the equation z=x+1+y.

$$z=\frac{1}{3}+\frac{1}{3}a+1-\frac{4}{3}+\frac{2}{3}a$$

Calculate z from z=\frac{1}{3}+\frac{1}{3}a+1-\frac{4}{3}+\frac{2}{3}a.

$$z=a$$

The system is now solved.

$$x=\frac{1}{3}+\frac{1}{3}a y=-\frac{4}{3}+\frac{2}{3}a z=a$$