Solve z=2-i/5+zika

$$z = \frac { 2 - i } { 5 + z i } k a$$

Solve for a

$$\left\{\begin{matrix}a=\frac{\left(\frac{2}{5}+\frac{1}{5}i\right)z\left(iz+5\right)}{k}\text{, }&k\neq 0\text{ and }z\neq 5i\\a\in \mathrm{C}\text{, }&k=0\text{ and }z=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by iz+5.

$$z\left(iz+5\right)=\left(2-i\right)ka$$

Use the distributive property to multiply z by iz+5.

$$iz^{2}+5z=\left(2-i\right)ka$$

Swap sides so that all variable terms are on the left hand side.

$$\left(2-i\right)ka=iz^{2}+5z$$

Divide both sides by \left(2-i\right)k.

$$\frac{\left(2-i\right)ka}{\left(2-i\right)k}=\frac{z\left(iz+5\right)}{\left(2-i\right)k}$$

Dividing by \left(2-i\right)k undoes the multiplication by \left(2-i\right)k.

$$a=\frac{z\left(iz+5\right)}{\left(2-i\right)k}$$

Divide z\left(5+iz\right) by \left(2-i\right)k.

$$a=\frac{\left(\frac{2}{5}+\frac{1}{5}i\right)z\left(iz+5\right)}{k}$$

Solve for k

$$\left\{\begin{matrix}k=\frac{\left(\frac{2}{5}+\frac{1}{5}i\right)z\left(iz+5\right)}{a}\text{, }&a\neq 0\text{ and }z\neq 5i\\k\in \mathrm{C}\text{, }&a=0\text{ and }z=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by iz+5.

$$z\left(iz+5\right)=\left(2-i\right)ka$$

Use the distributive property to multiply z by iz+5.

$$iz^{2}+5z=\left(2-i\right)ka$$

Swap sides so that all variable terms are on the left hand side.

$$\left(2-i\right)ka=iz^{2}+5z$$

The equation is in standard form.

$$\left(2-i\right)ak=iz^{2}+5z$$

Divide both sides by \left(2-i\right)a.

$$\frac{\left(2-i\right)ak}{\left(2-i\right)a}=\frac{z\left(iz+5\right)}{\left(2-i\right)a}$$

Dividing by \left(2-i\right)a undoes the multiplication by \left(2-i\right)a.

$$k=\frac{z\left(iz+5\right)}{\left(2-i\right)a}$$

Divide z\left(5+iz\right) by \left(2-i\right)a.

$$k=\frac{\left(\frac{2}{5}+\frac{1}{5}i\right)z\left(iz+5\right)}{a}$$