Solve {l}{2x-3y=4}{z=3y-x}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 2 x - 3 y = 4 }\\ { z = 3 y - x }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=a+4$$

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

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=3y-x 2x-3y=4 a=z$$

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

$$a=3y-x$$

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

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

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

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

Solve x=-a+3\left(-\frac{4}{3}+\frac{2}{3}x\right) for x.

$$x=4+a$$

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

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

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

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

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

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

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

$$z=a$$

The system is now solved.

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