Solve {l}{y+3x=6}{z=2x+3y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { y + 3 x = 6 }\\ { z = 2 x + 3 y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{18-a}{7}$$

$$y=\frac{3a-12}{7}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=2x+3y y+3x=6 a=z$$

Substitute 2x+3y for z in the equation a=z.

$$a=2x+3y$$

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

$$y=6-3x x=\frac{1}{2}a-\frac{3}{2}y$$

Substitute 6-3x for y in the equation x=\frac{1}{2}a-\frac{3}{2}y.

$$x=\frac{1}{2}a-\frac{3}{2}\left(6-3x\right)$$

Solve x=\frac{1}{2}a-\frac{3}{2}\left(6-3x\right) for x.

$$x=\frac{18}{7}-\frac{1}{7}a$$

Substitute \frac{18}{7}-\frac{1}{7}a for x in the equation y=6-3x.

$$y=6-3\left(\frac{18}{7}-\frac{1}{7}a\right)$$

Calculate y from y=6-3\left(\frac{18}{7}-\frac{1}{7}a\right).

$$y=-\frac{12}{7}+\frac{3}{7}a$$

Substitute -\frac{12}{7}+\frac{3}{7}a for y and \frac{18}{7}-\frac{1}{7}a for x in the equation z=2x+3y.

$$z=2\left(\frac{18}{7}-\frac{1}{7}a\right)+3\left(-\frac{12}{7}+\frac{3}{7}a\right)$$

Calculate z from z=2\left(\frac{18}{7}-\frac{1}{7}a\right)+3\left(-\frac{12}{7}+\frac{3}{7}a\right).

$$z=a$$

The system is now solved.

$$x=\frac{18}{7}-\frac{1}{7}a y=-\frac{12}{7}+\frac{3}{7}a z=a$$