Solve BC=BC+1/sqrt{3}

$$B C = \frac { B C + 1 } { \sqrt { 3 } }$$

Solve for B

$$B=\frac{\sqrt{3}+1}{2C}$$

$$C\neq 0$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{BC+1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$BC=\frac{\left(BC+1\right)\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$BC=\frac{\left(BC+1\right)\sqrt{3}}{3}$$

Use the distributive property to multiply BC+1 by \sqrt{3}.

$$BC=\frac{BC\sqrt{3}+\sqrt{3}}{3}$$

Subtract \frac{BC\sqrt{3}+\sqrt{3}}{3} from both sides.

$$BC-\frac{BC\sqrt{3}+\sqrt{3}}{3}=0$$

Multiply both sides of the equation by 3.

$$3BC-\left(BC\sqrt{3}+\sqrt{3}\right)=0$$

To find the opposite of BC\sqrt{3}+\sqrt{3}, find the opposite of each term.

$$3BC-BC\sqrt{3}-\sqrt{3}=0$$

Add \sqrt{3} to both sides. Anything plus zero gives itself.

$$3BC-BC\sqrt{3}=\sqrt{3}$$

Combine all terms containing B.

$$\left(3C-C\sqrt{3}\right)B=\sqrt{3}$$

The equation is in standard form.

$$\left(-\sqrt{3}C+3C\right)B=\sqrt{3}$$

Divide both sides by 3C-C\sqrt{3}.

$$\frac{\left(-\sqrt{3}C+3C\right)B}{-\sqrt{3}C+3C}=\frac{\sqrt{3}}{-\sqrt{3}C+3C}$$

Dividing by 3C-C\sqrt{3} undoes the multiplication by 3C-C\sqrt{3}.

$$B=\frac{\sqrt{3}}{-\sqrt{3}C+3C}$$

Divide \sqrt{3} by 3C-C\sqrt{3}.

$$B=\frac{\sqrt{3}+1}{2C}$$

Solve for C

$$C=\frac{\sqrt{3}+1}{2B}$$

$$B\neq 0$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{BC+1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$BC=\frac{\left(BC+1\right)\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$BC=\frac{\left(BC+1\right)\sqrt{3}}{3}$$

Use the distributive property to multiply BC+1 by \sqrt{3}.

$$BC=\frac{BC\sqrt{3}+\sqrt{3}}{3}$$

Subtract \frac{BC\sqrt{3}+\sqrt{3}}{3} from both sides.

$$BC-\frac{BC\sqrt{3}+\sqrt{3}}{3}=0$$

Multiply both sides of the equation by 3.

$$3BC-\left(BC\sqrt{3}+\sqrt{3}\right)=0$$

To find the opposite of BC\sqrt{3}+\sqrt{3}, find the opposite of each term.

$$3BC-BC\sqrt{3}-\sqrt{3}=0$$

Add \sqrt{3} to both sides. Anything plus zero gives itself.

$$3BC-BC\sqrt{3}=\sqrt{3}$$

Combine all terms containing C.

$$\left(3B-B\sqrt{3}\right)C=\sqrt{3}$$

The equation is in standard form.

$$\left(-\sqrt{3}B+3B\right)C=\sqrt{3}$$

Divide both sides by 3B-B\sqrt{3}.

$$\frac{\left(-\sqrt{3}B+3B\right)C}{-\sqrt{3}B+3B}=\frac{\sqrt{3}}{-\sqrt{3}B+3B}$$

Dividing by 3B-B\sqrt{3} undoes the multiplication by 3B-B\sqrt{3}.

$$C=\frac{\sqrt{3}}{-\sqrt{3}B+3B}$$

Divide \sqrt{3} by 3B-B\sqrt{3}.

$$C=\frac{\sqrt{3}+1}{2B}$$