Solve ax+3=bx-5c

$$a x + 3 = b x - 5 c$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{bx-5c-3}{x}\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&c=-\frac{3}{5}\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 3 from both sides.

$$ax=bx-5c-3$$

The equation is in standard form.

$$xa=bx-5c-3$$

Divide both sides by x.

$$\frac{xa}{x}=\frac{bx-5c-3}{x}$$

Dividing by x undoes the multiplication by x.

$$a=\frac{bx-5c-3}{x}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=\frac{ax+5c+3}{x}\text{, }&x\neq 0\\b\in \mathrm{C}\text{, }&c=-\frac{3}{5}\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$bx-5c=ax+3$$

Add 5c to both sides.

$$bx=ax+3+5c$$

The equation is in standard form.

$$xb=ax+5c+3$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{ax+5c+3}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{ax+5c+3}{x}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{bx-5c-3}{x}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&c=-\frac{3}{5}\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 3 from both sides.

$$ax=bx-5c-3$$

The equation is in standard form.

$$xa=bx-5c-3$$

Divide both sides by x.

$$\frac{xa}{x}=\frac{bx-5c-3}{x}$$

Dividing by x undoes the multiplication by x.

$$a=\frac{bx-5c-3}{x}$$

Solve for b

$$\left\{\begin{matrix}b=\frac{ax+5c+3}{x}\text{, }&x\neq 0\\b\in \mathrm{R}\text{, }&c=-\frac{3}{5}\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$bx-5c=ax+3$$

Add 5c to both sides.

$$bx=ax+3+5c$$

The equation is in standard form.

$$xb=ax+5c+3$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{ax+5c+3}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{ax+5c+3}{x}$$