Solve {l}{2x+y=1}{z=x+3y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 2 x + y = 1 }\\ { z = x + 3 y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

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

$$y=\frac{2a-1}{5}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=x+3y 2x+y=1 a=z$$

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

$$a=x+3y$$

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

$$y=-2x+1 x=a-3y$$

Substitute -2x+1 for y in the equation x=a-3y.

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

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

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

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

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

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

$$y=-\frac{1}{5}+\frac{2}{5}a$$

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

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

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

$$z=a$$

The system is now solved.

$$x=\frac{3}{5}-\frac{1}{5}a y=-\frac{1}{5}+\frac{2}{5}a z=a$$