Solve AB/20+BD=1/sqrt{3}

$$\frac { A B } { 20 + B D } = \frac { 1 } { \sqrt { 3 } }$$

Solve for A

$$A=\frac{\sqrt{3}\left(BD+20\right)}{3B}$$

$$B\neq 0\text{ and }\left(D=0\text{ or }B\neq -\frac{20}{D}\right)$$

Steps for Solving Linear Equation

Multiply both sides of the equation by BD+20.

$$AB=\frac{1}{3}\left(BD+20\right)\times 3^{\frac{1}{2}}$$

Use the distributive property to multiply \frac{1}{3} by BD+20.

$$AB=\left(\frac{1}{3}BD+\frac{20}{3}\right)\times 3^{\frac{1}{2}}$$

Use the distributive property to multiply \frac{1}{3}BD+\frac{20}{3} by 3^{\frac{1}{2}}.

$$AB=\frac{1}{3}BD\times 3^{\frac{1}{2}}+\frac{20}{3}\times 3^{\frac{1}{2}}$$

Reorder the terms.

$$AB=\frac{1}{3}\sqrt{3}BD+\frac{20}{3}\sqrt{3}$$

The equation is in standard form.

$$BA=\frac{\sqrt{3}BD+20\sqrt{3}}{3}$$

Divide both sides by B.

$$\frac{BA}{B}=\frac{\sqrt{3}\left(BD+20\right)}{3B}$$

Dividing by B undoes the multiplication by B.

$$A=\frac{\sqrt{3}\left(BD+20\right)}{3B}$$

Solve for B

$$B=\frac{20}{\sqrt{3}A-D}$$

$$A\neq 0\text{ and }A\neq \frac{\sqrt{3}D}{3}$$