Solve {l}{5x+4y=6}{z=9y+7x-11}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 5 x + 4 y = 6 }\\ { z = 9 y + 7 x - 11 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{10-4a}{17}$$

$$y=\frac{5a+13}{17}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=9y+7x-11 5x+4y=6 a=z$$

Substitute 9y+7x-11 for z in the equation a=z.

$$a=9y+7x-11$$

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

$$y=-\frac{5}{4}x+\frac{3}{2} x=\frac{11}{7}-\frac{9}{7}y+\frac{1}{7}a$$

Substitute -\frac{5}{4}x+\frac{3}{2} for y in the equation x=\frac{11}{7}-\frac{9}{7}y+\frac{1}{7}a.

$$x=\frac{11}{7}-\frac{9}{7}\left(-\frac{5}{4}x+\frac{3}{2}\right)+\frac{1}{7}a$$

Solve x=\frac{11}{7}-\frac{9}{7}\left(-\frac{5}{4}x+\frac{3}{2}\right)+\frac{1}{7}a for x.

$$x=\frac{10}{17}-\frac{4}{17}a$$

Substitute \frac{10}{17}-\frac{4}{17}a for x in the equation y=-\frac{5}{4}x+\frac{3}{2}.

$$y=-\frac{5}{4}\left(\frac{10}{17}-\frac{4}{17}a\right)+\frac{3}{2}$$

Calculate y from y=-\frac{5}{4}\left(\frac{10}{17}-\frac{4}{17}a\right)+\frac{3}{2}.

$$y=\frac{13}{17}+\frac{5}{17}a$$

Substitute \frac{13}{17}+\frac{5}{17}a for y and \frac{10}{17}-\frac{4}{17}a for x in the equation z=9y+7x-11.

$$z=9\left(\frac{13}{17}+\frac{5}{17}a\right)+7\left(\frac{10}{17}-\frac{4}{17}a\right)-11$$

Calculate z from z=9\left(\frac{13}{17}+\frac{5}{17}a\right)+7\left(\frac{10}{17}-\frac{4}{17}a\right)-11.

$$z=a$$

The system is now solved.

$$x=\frac{10}{17}-\frac{4}{17}a y=\frac{13}{17}+\frac{5}{17}a z=a$$