Solve {l}{x-y=9}{z=2x+5y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x - y = 9 }\\ { z = 2 x + 5 y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{a+45}{7}$$

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

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=2x+5y x-y=9 a=z$$

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

$$a=2x+5y$$

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

$$y=x-9 x=\frac{1}{2}a-\frac{5}{2}y$$

Substitute x-9 for y in the equation x=\frac{1}{2}a-\frac{5}{2}y.

$$x=\frac{1}{2}a-\frac{5}{2}\left(x-9\right)$$

Solve x=\frac{1}{2}a-\frac{5}{2}\left(x-9\right) for x.

$$x=\frac{45}{7}+\frac{1}{7}a$$

Substitute \frac{45}{7}+\frac{1}{7}a for x in the equation y=x-9.

$$y=\frac{45}{7}+\frac{1}{7}a-9$$

Calculate y from y=\frac{45}{7}+\frac{1}{7}a-9.

$$y=-\frac{18}{7}+\frac{1}{7}a$$

Substitute -\frac{18}{7}+\frac{1}{7}a for y and \frac{45}{7}+\frac{1}{7}a for x in the equation z=2x+5y.

$$z=2\left(\frac{45}{7}+\frac{1}{7}a\right)+5\left(-\frac{18}{7}+\frac{1}{7}a\right)$$

Calculate z from z=2\left(\frac{45}{7}+\frac{1}{7}a\right)+5\left(-\frac{18}{7}+\frac{1}{7}a\right).

$$z=a$$

The system is now solved.

$$x=\frac{45}{7}+\frac{1}{7}a y=-\frac{18}{7}+\frac{1}{7}a z=a$$