Solve {l}{x+4y=20}{z=x+12y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x + 4 y = 20 }\\ { z = x + 12 y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=-\frac{a}{2}+30$$

$$y=\frac{a}{8}-\frac{5}{2}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=x+12y x+4y=20 a=z$$

Substitute x+12y for z in the equation a=z.

$$a=x+12y$$

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

$$y=-\frac{1}{4}x+5 x=a-12y$$

Substitute -\frac{1}{4}x+5 for y in the equation x=a-12y.

$$x=a-12\left(-\frac{1}{4}x+5\right)$$

Solve x=a-12\left(-\frac{1}{4}x+5\right) for x.

$$x=30-\frac{1}{2}a$$

Substitute 30-\frac{1}{2}a for x in the equation y=-\frac{1}{4}x+5.

$$y=-\frac{1}{4}\left(30-\frac{1}{2}a\right)+5$$

Calculate y from y=-\frac{1}{4}\left(30-\frac{1}{2}a\right)+5.

$$y=-\frac{5}{2}+\frac{1}{8}a$$

Substitute -\frac{5}{2}+\frac{1}{8}a for y and 30-\frac{1}{2}a for x in the equation z=x+12y.

$$z=30-\frac{1}{2}a+12\left(-\frac{5}{2}+\frac{1}{8}a\right)$$

Calculate z from z=30-\frac{1}{2}a+12\left(-\frac{5}{2}+\frac{1}{8}a\right).

$$z=a$$

The system is now solved.

$$x=30-\frac{1}{2}a y=-\frac{5}{2}+\frac{1}{8}a z=a$$