Solve {l}{3x-y+30=0}{z=12x-5y+240}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 3 x - y + 30 = 0 }\\ { z = 12 x - 5 y + 240 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=-\frac{a}{3}+30$$

$$y=120-a$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=12x-5y+240 3x-y+30=0 a=z$$

Substitute 12x-5y+240 for z in the equation a=z.

$$a=12x-5y+240$$

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

$$y=30+3x x=\frac{5}{12}y+\frac{1}{12}a-20$$

Substitute 30+3x for y in the equation x=\frac{5}{12}y+\frac{1}{12}a-20.

$$x=\frac{5}{12}\left(30+3x\right)+\frac{1}{12}a-20$$

Solve x=\frac{5}{12}\left(30+3x\right)+\frac{1}{12}a-20 for x.

$$x=30-\frac{1}{3}a$$

Substitute 30-\frac{1}{3}a for x in the equation y=30+3x.

$$y=30+3\left(30-\frac{1}{3}a\right)$$

Calculate y from y=30+3\left(30-\frac{1}{3}a\right).

$$y=120-a$$

Substitute 120-a for y and 30-\frac{1}{3}a for x in the equation z=12x-5y+240.

$$z=12\left(30-\frac{1}{3}a\right)-5\left(120-a\right)+240$$

Calculate z from z=12\left(30-\frac{1}{3}a\right)-5\left(120-a\right)+240.

$$z=a$$

The system is now solved.

$$x=30-\frac{1}{3}a y=120-a z=a$$