Solve fquadz_{1}=2+3iquadar

$$f \quad z _ { 1 } = 2 + 3 i \quad ar$$

Solve for a

$$\left\{\begin{matrix}a=\frac{i\left(2-fz_{1}\right)}{3r}\text{, }&r\neq 0\\a\in \mathrm{C}\text{, }&f=\frac{2}{z_{1}}\text{ and }z_{1}\neq 0\text{ and }r=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$2+3iar=fz_{1}$$

Subtract 2 from both sides.

$$3iar=fz_{1}-2$$

The equation is in standard form.

$$3ira=fz_{1}-2$$

Divide both sides by 3ir.

$$\frac{3ira}{3ir}=\frac{fz_{1}-2}{3ir}$$

Dividing by 3ir undoes the multiplication by 3ir.

$$a=\frac{fz_{1}-2}{3ir}$$

Divide fz_{1}-2 by 3ir.

$$a=-\frac{i\left(fz_{1}-2\right)}{3r}$$

Solve for f

$$\left\{\begin{matrix}f=\frac{3iar+2}{z_{1}}\text{, }&z_{1}\neq 0\\f\in \mathrm{C}\text{, }&a=\frac{2i}{3r}\text{ and }r\neq 0\text{ and }z_{1}=0\end{matrix}\right.$$

Steps for Solving Linear Equation

The equation is in standard form.

$$z_{1}f=3iar+2$$

Divide both sides by z_{1}.

$$\frac{z_{1}f}{z_{1}}=\frac{3iar+2}{z_{1}}$$

Dividing by z_{1} undoes the multiplication by z_{1}.

$$f=\frac{3iar+2}{z_{1}}$$