Solve {l}{6x+3y=28}{z=x+5y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 6 x + 3 y = 28 }\\ { z = x + 5 y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=-\frac{a}{9}+\frac{140}{27}$$

$$y=\frac{2a}{9}-\frac{28}{27}$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=x+5y 6x+3y=28 a=z$$

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

$$a=x+5y$$

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

$$y=\frac{28}{3}-2x x=a-5y$$

Substitute \frac{28}{3}-2x for y in the equation x=a-5y.

$$x=a-5\left(\frac{28}{3}-2x\right)$$

Solve x=a-5\left(\frac{28}{3}-2x\right) for x.

$$x=\frac{140}{27}-\frac{1}{9}a$$

Substitute \frac{140}{27}-\frac{1}{9}a for x in the equation y=\frac{28}{3}-2x.

$$y=\frac{28}{3}-2\left(\frac{140}{27}-\frac{1}{9}a\right)$$

Calculate y from y=\frac{28}{3}-2\left(\frac{140}{27}-\frac{1}{9}a\right).

$$y=-\frac{28}{27}+\frac{2}{9}a$$

Substitute -\frac{28}{27}+\frac{2}{9}a for y and \frac{140}{27}-\frac{1}{9}a for x in the equation z=x+5y.

$$z=\frac{140}{27}-\frac{1}{9}a+5\left(-\frac{28}{27}+\frac{2}{9}a\right)$$

Calculate z from z=\frac{140}{27}-\frac{1}{9}a+5\left(-\frac{28}{27}+\frac{2}{9}a\right).

$$z=a$$

The system is now solved.

$$x=\frac{140}{27}-\frac{1}{9}a y=-\frac{28}{27}+\frac{2}{9}a z=a$$