Solve {l}{x+y=10}{z=2x-3y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x + y = 10 }\\ { z = 2 x - 3 y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

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

$$y=-\frac{a}{5}+4$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=2x-3y x+y=10 a=z$$

Substitute 2x-3y for z in the equation a=z.

$$a=2x-3y$$

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

$$y=-x+10 x=\frac{1}{2}a+\frac{3}{2}y$$

Substitute -x+10 for y in the equation x=\frac{1}{2}a+\frac{3}{2}y.

$$x=\frac{1}{2}a+\frac{3}{2}\left(-x+10\right)$$

Solve x=\frac{1}{2}a+\frac{3}{2}\left(-x+10\right) for x.

$$x=6+\frac{1}{5}a$$

Substitute 6+\frac{1}{5}a for x in the equation y=-x+10.

$$y=-\left(6+\frac{1}{5}a\right)+10$$

Calculate y from y=-\left(6+\frac{1}{5}a\right)+10.

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

Substitute 4-\frac{1}{5}a for y and 6+\frac{1}{5}a for x in the equation z=2x-3y.

$$z=2\left(6+\frac{1}{5}a\right)-3\left(4-\frac{1}{5}a\right)$$

Calculate z from z=2\left(6+\frac{1}{5}a\right)-3\left(4-\frac{1}{5}a\right).

$$z=a$$

The system is now solved.

$$x=6+\frac{1}{5}a y=4-\frac{1}{5}a z=a$$