Solve {l}{5y+x=1}{z=x+3y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 5 y + x = 1 }\\ { z = x + 3 y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{5a-3}{2}$$

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

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=x+3y 5y+x=1 a=z$$

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.

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

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

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

Solve x=a-3\left(-\frac{1}{5}x+\frac{1}{5}\right) for x.

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

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

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

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

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

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

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

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

$$z=a$$

The system is now solved.

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