Solve d_{1}quadf(x)=2x+3

$$d _ { 1 } \quad f ( x ) = 2 x + 3$$

Solve for d_1 (complex solution)

$$\left\{\begin{matrix}d_{1}=\frac{2x+3}{fx}\text{, }&x\neq 0\text{ and }f\neq 0\\d_{1}\in \mathrm{C}\text{, }&x=-\frac{3}{2}\text{ and }f=0\end{matrix}\right.$$

Steps for Solving Linear Equation

The equation is in standard form.

$$fxd_{1}=2x+3$$

Divide both sides by fx.

$$\frac{fxd_{1}}{fx}=\frac{2x+3}{fx}$$

Dividing by fx undoes the multiplication by fx.

$$d_{1}=\frac{2x+3}{fx}$$

Solve for f (complex solution)

$$\left\{\begin{matrix}f=\frac{2x+3}{d_{1}x}\text{, }&x\neq 0\text{ and }d_{1}\neq 0\\f\in \mathrm{C}\text{, }&x=-\frac{3}{2}\text{ and }d_{1}=0\end{matrix}\right.$$

Steps for Solving Linear Equation

The equation is in standard form.

$$d_{1}xf=2x+3$$

Divide both sides by d_{1}x.

$$\frac{d_{1}xf}{d_{1}x}=\frac{2x+3}{d_{1}x}$$

Dividing by d_{1}x undoes the multiplication by d_{1}x.

$$f=\frac{2x+3}{d_{1}x}$$

Solve for d_1

$$\left\{\begin{matrix}d_{1}=\frac{2x+3}{fx}\text{, }&x\neq 0\text{ and }f\neq 0\\d_{1}\in \mathrm{R}\text{, }&x=-\frac{3}{2}\text{ and }f=0\end{matrix}\right.$$

Steps for Solving Linear Equation

The equation is in standard form.

$$fxd_{1}=2x+3$$

Divide both sides by fx.

$$\frac{fxd_{1}}{fx}=\frac{2x+3}{fx}$$

Dividing by fx undoes the multiplication by fx.

$$d_{1}=\frac{2x+3}{fx}$$

Solve for f

$$\left\{\begin{matrix}f=\frac{2x+3}{d_{1}x}\text{, }&x\neq 0\text{ and }d_{1}\neq 0\\f\in \mathrm{R}\text{, }&x=-\frac{3}{2}\text{ and }d_{1}=0\end{matrix}\right.$$

Steps for Solving Linear Equation

The equation is in standard form.

$$d_{1}xf=2x+3$$

Divide both sides by d_{1}x.

$$\frac{d_{1}xf}{d_{1}x}=\frac{2x+3}{d_{1}x}$$

Dividing by d_{1}x undoes the multiplication by d_{1}x.

$$f=\frac{2x+3}{d_{1}x}$$