Solve {l}{x+y=4}{z=y-2x+1}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x + y = 4 }\\ { z = y - 2 x + 1 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{5-a}{3}$$

$$y=\frac{a+7}{3}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=y-2x+1 x+y=4 a=z$$

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

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

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

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

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

$$x=\frac{1}{2}+\frac{1}{2}\left(-x+4\right)-\frac{1}{2}a$$

Solve x=\frac{1}{2}+\frac{1}{2}\left(-x+4\right)-\frac{1}{2}a for x.

$$x=\frac{5}{3}-\frac{1}{3}a$$

Substitute \frac{5}{3}-\frac{1}{3}a for x in the equation y=-x+4.

$$y=-\left(\frac{5}{3}-\frac{1}{3}a\right)+4$$

Calculate y from y=-\left(\frac{5}{3}-\frac{1}{3}a\right)+4.

$$y=\frac{7}{3}+\frac{1}{3}a$$

Substitute \frac{7}{3}+\frac{1}{3}a for y and \frac{5}{3}-\frac{1}{3}a for x in the equation z=y-2x+1.

$$z=\frac{7}{3}+\frac{1}{3}a-2\left(\frac{5}{3}-\frac{1}{3}a\right)+1$$

Calculate z from z=\frac{7}{3}+\frac{1}{3}a-2\left(\frac{5}{3}-\frac{1}{3}a\right)+1.

$$z=a$$

The system is now solved.

$$x=\frac{5}{3}-\frac{1}{3}a y=\frac{7}{3}+\frac{1}{3}a z=a$$