Solve {l}{x=3x-y}{z=3x+4}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x = 3 x - y }\\ { z = 3 x + 4 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{a-4}{3}$$

$$y=\frac{2\left(a-4\right)}{3}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=3x+4 x=3x-y a=z$$

Substitute 3x+4 for z in the equation a=z.

$$a=3x+4$$

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

$$y=2x x=-\frac{4}{3}+\frac{1}{3}a$$

Substitute -\frac{4}{3}+\frac{1}{3}a for x in the equation y=2x.

$$y=2\left(-\frac{4}{3}+\frac{1}{3}a\right)$$

Calculate y from y=2\left(-\frac{4}{3}+\frac{1}{3}a\right).

$$y=-\frac{8}{3}+\frac{2}{3}a$$

Substitute -\frac{4}{3}+\frac{1}{3}a for x in the equation z=3x+4.

$$z=3\left(-\frac{4}{3}+\frac{1}{3}a\right)+4$$

Calculate z from z=3\left(-\frac{4}{3}+\frac{1}{3}a\right)+4.

$$z=a$$

The system is now solved.

$$x=-\frac{4}{3}+\frac{1}{3}a y=-\frac{8}{3}+\frac{2}{3}a z=a$$