Solve {l}{x+y=3}{z=2x+5y-4}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x + y = 3 }\\ { z = 2 x + 5 y - 4 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{11-a}{3}$$

$$y=\frac{a-2}{3}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=2x+5y-4 x+y=3 a=z$$

Substitute 2x+5y-4 for z in the equation a=z.

$$a=2x+5y-4$$

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

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

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

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

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

$$x=\frac{11}{3}-\frac{1}{3}a$$

Substitute \frac{11}{3}-\frac{1}{3}a for x in the equation y=-x+3.

$$y=-\left(\frac{11}{3}-\frac{1}{3}a\right)+3$$

Calculate y from y=-\left(\frac{11}{3}-\frac{1}{3}a\right)+3.

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

Substitute -\frac{2}{3}+\frac{1}{3}a for y and \frac{11}{3}-\frac{1}{3}a for x in the equation z=2x+5y-4.

$$z=2\left(\frac{11}{3}-\frac{1}{3}a\right)+5\left(-\frac{2}{3}+\frac{1}{3}a\right)-4$$

Calculate z from z=2\left(\frac{11}{3}-\frac{1}{3}a\right)+5\left(-\frac{2}{3}+\frac{1}{3}a\right)-4.

$$z=a$$

The system is now solved.

$$x=\frac{11}{3}-\frac{1}{3}a y=-\frac{2}{3}+\frac{1}{3}a z=a$$