Solve {l}{4x-y=19}{z={(5y-2x)}}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 4 x - y = 19 }\\ { z = {(5 y - 2 x)} }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{a+95}{18}$$

$$y=\frac{2a+19}{9}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=5y-2x 4x-y=19 a=z$$

Substitute 5y-2x for z in the equation a=z.

$$a=5y-2x$$

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

$$y=4x-19 x=-\frac{1}{2}a+\frac{5}{2}y$$

Substitute 4x-19 for y in the equation x=-\frac{1}{2}a+\frac{5}{2}y.

$$x=-\frac{1}{2}a+\frac{5}{2}\left(4x-19\right)$$

Solve x=-\frac{1}{2}a+\frac{5}{2}\left(4x-19\right) for x.

$$x=\frac{95}{18}+\frac{1}{18}a$$

Substitute \frac{95}{18}+\frac{1}{18}a for x in the equation y=4x-19.

$$y=4\left(\frac{95}{18}+\frac{1}{18}a\right)-19$$

Calculate y from y=4\left(\frac{95}{18}+\frac{1}{18}a\right)-19.

$$y=\frac{19}{9}+\frac{2}{9}a$$

Substitute \frac{19}{9}+\frac{2}{9}a for y and \frac{95}{18}+\frac{1}{18}a for x in the equation z=5y-2x.

$$z=5\left(\frac{19}{9}+\frac{2}{9}a\right)-2\left(\frac{95}{18}+\frac{1}{18}a\right)$$

Calculate z from z=5\left(\frac{19}{9}+\frac{2}{9}a\right)-2\left(\frac{95}{18}+\frac{1}{18}a\right).

$$z=a$$

The system is now solved.

$$x=\frac{95}{18}+\frac{1}{18}a y=\frac{19}{9}+\frac{2}{9}a z=a$$