Solve {l}{3x-4y=10}{z=4x+3y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 3 x - 4 y = 10 }\\ { z = 4 x + 3 y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{4a}{25}+\frac{6}{5}$$

$$y=\frac{3a}{25}-\frac{8}{5}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=4x+3y 3x-4y=10 a=z$$

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

$$a=4x+3y$$

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

$$y=\frac{3}{4}x-\frac{5}{2} x=\frac{1}{4}a-\frac{3}{4}y$$

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

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

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

$$x=\frac{6}{5}+\frac{4}{25}a$$

Substitute \frac{6}{5}+\frac{4}{25}a for x in the equation y=\frac{3}{4}x-\frac{5}{2}.

$$y=\frac{3}{4}\left(\frac{6}{5}+\frac{4}{25}a\right)-\frac{5}{2}$$

Calculate y from y=\frac{3}{4}\left(\frac{6}{5}+\frac{4}{25}a\right)-\frac{5}{2}.

$$y=-\frac{8}{5}+\frac{3}{25}a$$

Substitute -\frac{8}{5}+\frac{3}{25}a for y and \frac{6}{5}+\frac{4}{25}a for x in the equation z=4x+3y.

$$z=4\left(\frac{6}{5}+\frac{4}{25}a\right)+3\left(-\frac{8}{5}+\frac{3}{25}a\right)$$

Calculate z from z=4\left(\frac{6}{5}+\frac{4}{25}a\right)+3\left(-\frac{8}{5}+\frac{3}{25}a\right).

$$z=a$$

The system is now solved.

$$x=\frac{6}{5}+\frac{4}{25}a y=-\frac{8}{5}+\frac{3}{25}a z=a$$