Solve Ay=5(y-B)C

$$A y = 5 ( y - B ) C$$

Solve for A (complex solution)

$$\left\{\begin{matrix}A=\frac{5C\left(y-B\right)}{y}\text{, }&y\neq 0\\A\in \mathrm{C}\text{, }&\left(B=0\text{ or }C=0\right)\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 5 by y-B.

$$Ay=\left(5y-5B\right)C$$

Use the distributive property to multiply 5y-5B by C.

$$Ay=5yC-5BC$$

The equation is in standard form.

$$yA=5Cy-5BC$$

Divide both sides by y.

$$\frac{yA}{y}=\frac{5C\left(y-B\right)}{y}$$

Dividing by y undoes the multiplication by y.

$$A=\frac{5C\left(y-B\right)}{y}$$

Solve for B (complex solution)

$$\left\{\begin{matrix}B=-\frac{Ay}{5C}+y\text{, }&C\neq 0\\B\in \mathrm{C}\text{, }&\left(A=0\text{ or }y=0\right)\text{ and }C=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 5 by y-B.

$$Ay=\left(5y-5B\right)C$$

Use the distributive property to multiply 5y-5B by C.

$$Ay=5yC-5BC$$

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

$$5yC-5BC=Ay$$

Subtract 5yC from both sides.

$$-5BC=Ay-5yC$$

The equation is in standard form.

$$\left(-5C\right)B=Ay-5Cy$$

Divide both sides by -5C.

$$\frac{\left(-5C\right)B}{-5C}=\frac{y\left(A-5C\right)}{-5C}$$

Dividing by -5C undoes the multiplication by -5C.

$$B=\frac{y\left(A-5C\right)}{-5C}$$

Divide y\left(A-5C\right) by -5C.

$$B=-\frac{Ay}{5C}+y$$

Solve for A

$$\left\{\begin{matrix}A=\frac{5C\left(y-B\right)}{y}\text{, }&y\neq 0\\A\in \mathrm{R}\text{, }&\left(B=0\text{ or }C=0\right)\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 5 by y-B.

$$Ay=\left(5y-5B\right)C$$

Use the distributive property to multiply 5y-5B by C.

$$Ay=5yC-5BC$$

The equation is in standard form.

$$yA=5Cy-5BC$$

Divide both sides by y.

$$\frac{yA}{y}=\frac{5C\left(y-B\right)}{y}$$

Dividing by y undoes the multiplication by y.

$$A=\frac{5C\left(y-B\right)}{y}$$

Solve for B

$$\left\{\begin{matrix}B=-\frac{Ay}{5C}+y\text{, }&C\neq 0\\B\in \mathrm{R}\text{, }&\left(A=0\text{ or }y=0\right)\text{ and }C=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 5 by y-B.

$$Ay=\left(5y-5B\right)C$$

Use the distributive property to multiply 5y-5B by C.

$$Ay=5yC-5BC$$

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

$$5yC-5BC=Ay$$

Subtract 5yC from both sides.

$$-5BC=Ay-5yC$$

The equation is in standard form.

$$\left(-5C\right)B=Ay-5Cy$$

Divide both sides by -5C.

$$\frac{\left(-5C\right)B}{-5C}=\frac{y\left(A-5C\right)}{-5C}$$

Dividing by -5C undoes the multiplication by -5C.

$$B=\frac{y\left(A-5C\right)}{-5C}$$

Divide y\left(A-5C\right) by -5C.

$$B=-\frac{Ay}{5C}+y$$