Solve {l}{3x-y-6i=2+xi-2yi}{z=x+y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 3 x - y - 6 i = 2 + x i - 2 y i }\\ { z = x + y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z

$$x=\left(\frac{2}{5}-\frac{3}{10}i\right)a+\left(\frac{1}{5}+\frac{1}{10}i\right)\xi +\left(-\frac{1}{5}+\frac{7}{5}i\right)$$

$$y=\left(\frac{3}{5}+\frac{3}{10}i\right)a+\left(-\frac{1}{5}-\frac{1}{10}i\right)\xi +\left(\frac{1}{5}-\frac{7}{5}i\right)$$

$$z=a$$

Short Steps Using Substitution

Reorder the equations.

$$z=x+y 3x-y-6i=2+\xi -2yi a=z$$

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

$$a=x+y$$

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

$$y=\left(\frac{3}{5}+\frac{6}{5}i\right)x+\left(2-2i\right)+\left(-\frac{1}{5}-\frac{2}{5}i\right)\xi x=a-y$$

Substitute \left(\frac{3}{5}+\frac{6}{5}i\right)x+\left(2-2i\right)+\left(-\frac{1}{5}-\frac{2}{5}i\right)\xi for y in the equation x=a-y.

$$x=a-\left(\left(\frac{3}{5}+\frac{6}{5}i\right)x+\left(2-2i\right)+\left(-\frac{1}{5}-\frac{2}{5}i\right)\xi \right)$$

Solve x=a-\left(\left(\frac{3}{5}+\frac{6}{5}i\right)x+\left(2-2i\right)+\left(-\frac{1}{5}-\frac{2}{5}i\right)\xi \right) for x.

$$x=-\frac{1}{5}+\frac{7}{5}i+\left(\frac{1}{5}+\frac{1}{10}i\right)\xi +\left(\frac{2}{5}-\frac{3}{10}i\right)a$$

Substitute -\frac{1}{5}+\frac{7}{5}i+\left(\frac{1}{5}+\frac{1}{10}i\right)\xi +\left(\frac{2}{5}-\frac{3}{10}i\right)a for x in the equation y=\left(\frac{3}{5}+\frac{6}{5}i\right)x+\left(2-2i\right)+\left(-\frac{1}{5}-\frac{2}{5}i\right)\xi .

$$y=\left(\frac{3}{5}+\frac{6}{5}i\right)\left(-\frac{1}{5}+\frac{7}{5}i+\left(\frac{1}{5}+\frac{1}{10}i\right)\xi +\left(\frac{2}{5}-\frac{3}{10}i\right)a\right)+\left(2-2i\right)+\left(-\frac{1}{5}-\frac{2}{5}i\right)\xi$$

Calculate y from y=\left(\frac{3}{5}+\frac{6}{5}i\right)\left(-\frac{1}{5}+\frac{7}{5}i+\left(\frac{1}{5}+\frac{1}{10}i\right)\xi +\left(\frac{2}{5}-\frac{3}{10}i\right)a\right)+\left(2-2i\right)+\left(-\frac{1}{5}-\frac{2}{5}i\right)\xi .

$$y=\frac{1}{5}-\frac{7}{5}i+\left(-\frac{1}{5}-\frac{1}{10}i\right)\xi +\left(\frac{3}{5}+\frac{3}{10}i\right)a$$

Substitute \frac{1}{5}-\frac{7}{5}i+\left(-\frac{1}{5}-\frac{1}{10}i\right)\xi +\left(\frac{3}{5}+\frac{3}{10}i\right)a for y and -\frac{1}{5}+\frac{7}{5}i+\left(\frac{1}{5}+\frac{1}{10}i\right)\xi +\left(\frac{2}{5}-\frac{3}{10}i\right)a for x in the equation z=x+y.

$$z=-\frac{1}{5}+\frac{7}{5}i+\left(\frac{1}{5}+\frac{1}{10}i\right)\xi +\left(\frac{2}{5}-\frac{3}{10}i\right)a+\left(\frac{1}{5}-\frac{7}{5}i\right)+\left(-\frac{1}{5}-\frac{1}{10}i\right)\xi +\left(\frac{3}{5}+\frac{3}{10}i\right)a$$

Calculate z from z=-\frac{1}{5}+\frac{7}{5}i+\left(\frac{1}{5}+\frac{1}{10}i\right)\xi +\left(\frac{2}{5}-\frac{3}{10}i\right)a+\left(\frac{1}{5}-\frac{7}{5}i\right)+\left(-\frac{1}{5}-\frac{1}{10}i\right)\xi +\left(\frac{3}{5}+\frac{3}{10}i\right)a.

$$z=a$$

The system is now solved.

$$x=-\frac{1}{5}+\frac{7}{5}i+\left(\frac{1}{5}+\frac{1}{10}i\right)\xi +\left(\frac{2}{5}-\frac{3}{10}i\right)a y=\frac{1}{5}-\frac{7}{5}i+\left(-\frac{1}{5}-\frac{1}{10}i\right)\xi +\left(\frac{3}{5}+\frac{3}{10}i\right)a z=a$$