Solve {l}{x+y=2}{z=3x+y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x + y = 2 }\\ { z = 3 x + y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

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

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

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=3x+y x+y=2 a=z$$

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

$$a=3x+y$$

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

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

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

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

Solve x=\frac{1}{3}a-\frac{1}{3}\left(-x+2\right) for x.

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

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

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

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

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

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

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

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

$$z=a$$

The system is now solved.

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