Solve {l}{x+y=1}{z=4x+7y-28}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x + y = 1 }\\ { z = 4 x + 7 y - 28 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=-\frac{a}{3}-7$$

$$y=\frac{a}{3}+8$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=4x+7y-28 x+y=1 a=z$$

Substitute 4x+7y-28 for z in the equation a=z.

$$a=4x+7y-28$$

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

$$y=-x+1 x=-\frac{7}{4}y+\frac{1}{4}a+7$$

Substitute -x+1 for y in the equation x=-\frac{7}{4}y+\frac{1}{4}a+7.

$$x=-\frac{7}{4}\left(-x+1\right)+\frac{1}{4}a+7$$

Solve x=-\frac{7}{4}\left(-x+1\right)+\frac{1}{4}a+7 for x.

$$x=-7-\frac{1}{3}a$$

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

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

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

$$y=8+\frac{1}{3}a$$

Substitute 8+\frac{1}{3}a for y and -7-\frac{1}{3}a for x in the equation z=4x+7y-28.

$$z=4\left(-7-\frac{1}{3}a\right)+7\left(8+\frac{1}{3}a\right)-28$$

Calculate z from z=4\left(-7-\frac{1}{3}a\right)+7\left(8+\frac{1}{3}a\right)-28.

$$z=a$$

The system is now solved.

$$x=-7-\frac{1}{3}a y=8+\frac{1}{3}a z=a$$