Solve {l}{x=2y-3}{z=8y-x+11}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x = 2 y - 3 }\\ { z = 8 y - x + 11 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{a-23}{3}$$

$$y=\frac{a-14}{6}$$

$$z=a$$

Short Steps Using Substitution

Substitute 2y-3 for x in the equation z=8y-x+11.

$$z=8y-\left(2y-3\right)+11$$

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

$$y=-\frac{7}{3}+\frac{1}{6}z z=a$$

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

$$y=-\frac{7}{3}+\frac{1}{6}a$$

Substitute -\frac{7}{3}+\frac{1}{6}a for y in the equation x=2y-3.

$$x=2\left(-\frac{7}{3}+\frac{1}{6}a\right)-3$$

Calculate x from x=2\left(-\frac{7}{3}+\frac{1}{6}a\right)-3.

$$x=-\frac{23}{3}+\frac{1}{3}a$$

The system is now solved.

$$x=-\frac{23}{3}+\frac{1}{3}a y=-\frac{7}{3}+\frac{1}{6}a z=a$$