Solve x=abc+y

$$x = abc+y$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{y-x}{bc}\text{, }&c\neq 0\text{ and }b\neq 0\\a\in \mathrm{C}\text{, }&x=y\text{ and }\left(c=0\text{ or }b=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$abc+y=x$$

Subtract y from both sides.

$$abc=x-y$$

The equation is in standard form.

$$bca=x-y$$

Divide both sides by bc.

$$\frac{bca}{bc}=\frac{x-y}{bc}$$

Dividing by bc undoes the multiplication by bc.

$$a=\frac{x-y}{bc}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-\frac{y-x}{ac}\text{, }&c\neq 0\text{ and }a\neq 0\\b\in \mathrm{C}\text{, }&x=y\text{ and }\left(c=0\text{ or }a=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$abc+y=x$$

Subtract y from both sides.

$$abc=x-y$$

The equation is in standard form.

$$acb=x-y$$

Divide both sides by ac.

$$\frac{acb}{ac}=\frac{x-y}{ac}$$

Dividing by ac undoes the multiplication by ac.

$$b=\frac{x-y}{ac}$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{y-x}{bc}\text{, }&c\neq 0\text{ and }b\neq 0\\a\in \mathrm{R}\text{, }&x=y\text{ and }\left(c=0\text{ or }b=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$abc+y=x$$

Subtract y from both sides.

$$abc=x-y$$

The equation is in standard form.

$$bca=x-y$$

Divide both sides by bc.

$$\frac{bca}{bc}=\frac{x-y}{bc}$$

Dividing by bc undoes the multiplication by bc.

$$a=\frac{x-y}{bc}$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{y-x}{ac}\text{, }&c\neq 0\text{ and }a\neq 0\\b\in \mathrm{R}\text{, }&x=y\text{ and }\left(c=0\text{ or }a=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$abc+y=x$$

Subtract y from both sides.

$$abc=x-y$$

The equation is in standard form.

$$acb=x-y$$

Divide both sides by ac.

$$\frac{acb}{ac}=\frac{x-y}{ac}$$

Dividing by ac undoes the multiplication by ac.

$$b=\frac{x-y}{ac}$$