Solve {l}{-x+y=1}{z=8x+y-3}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { -x + y = 1 }\\ { z = 8 x + y - 3 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{a+2}{9}$$

$$y=\frac{a+11}{9}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=8x+y-3 -x+y=1 a=z$$

Substitute 8x+y-3 for z in the equation a=z.

$$a=8x+y-3$$

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

$$y=x+1 x=\frac{3}{8}-\frac{1}{8}y+\frac{1}{8}a$$

Substitute x+1 for y in the equation x=\frac{3}{8}-\frac{1}{8}y+\frac{1}{8}a.

$$x=\frac{3}{8}-\frac{1}{8}\left(x+1\right)+\frac{1}{8}a$$

Solve x=\frac{3}{8}-\frac{1}{8}\left(x+1\right)+\frac{1}{8}a for x.

$$x=\frac{2}{9}+\frac{1}{9}a$$

Substitute \frac{2}{9}+\frac{1}{9}a for x in the equation y=x+1.

$$y=\frac{2}{9}+\frac{1}{9}a+1$$

Calculate y from y=\frac{2}{9}+\frac{1}{9}a+1.

$$y=\frac{11}{9}+\frac{1}{9}a$$

Substitute \frac{11}{9}+\frac{1}{9}a for y and \frac{2}{9}+\frac{1}{9}a for x in the equation z=8x+y-3.

$$z=8\left(\frac{2}{9}+\frac{1}{9}a\right)+\frac{11}{9}+\frac{1}{9}a-3$$

Calculate z from z=8\left(\frac{2}{9}+\frac{1}{9}a\right)+\frac{11}{9}+\frac{1}{9}a-3.

$$z=a$$

The system is now solved.

$$x=\frac{2}{9}+\frac{1}{9}a y=\frac{11}{9}+\frac{1}{9}a z=a$$