Solve D=frac{v_{0}}{sqrt{2}+sqrt{3}}

$$D = \frac { v _ { 0 } } { \sqrt { 2 } + \sqrt { 3 } }$$

Solve for v_0

$$v_{0}=\left(\sqrt{2}+\sqrt{3}\right)D$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{v_{0}}{\sqrt{2}+\sqrt{3}} by multiplying numerator and denominator by \sqrt{2}-\sqrt{3}.

$$D=\frac{v_{0}\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}$$

Consider \left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$D=\frac{v_{0}\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

Square \sqrt{2}. Square \sqrt{3}.

$$D=\frac{v_{0}\left(\sqrt{2}-\sqrt{3}\right)}{2-3}$$

Subtract 3 from 2 to get -1.

$$D=\frac{v_{0}\left(\sqrt{2}-\sqrt{3}\right)}{-1}$$

Anything divided by -1 gives its opposite.

$$D=-v_{0}\left(\sqrt{2}-\sqrt{3}\right)$$

Use the distributive property to multiply v_{0} by \sqrt{2}-\sqrt{3}.

$$D=-\left(v_{0}\sqrt{2}-v_{0}\sqrt{3}\right)$$

To find the opposite of v_{0}\sqrt{2}-v_{0}\sqrt{3}, find the opposite of each term.

$$D=-v_{0}\sqrt{2}+v_{0}\sqrt{3}$$

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

$$-v_{0}\sqrt{2}+v_{0}\sqrt{3}=D$$

Combine all terms containing v_{0}.

$$\left(-\sqrt{2}+\sqrt{3}\right)v_{0}=D$$

The equation is in standard form.

$$\left(\sqrt{3}-\sqrt{2}\right)v_{0}=D$$

Divide both sides by -\sqrt{2}+\sqrt{3}.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)v_{0}}{\sqrt{3}-\sqrt{2}}=\frac{D}{\sqrt{3}-\sqrt{2}}$$

Dividing by -\sqrt{2}+\sqrt{3} undoes the multiplication by -\sqrt{2}+\sqrt{3}.

$$v_{0}=\frac{D}{\sqrt{3}-\sqrt{2}}$$

Divide D by -\sqrt{2}+\sqrt{3}.

$$v_{0}=\left(\sqrt{2}+\sqrt{3}\right)D$$

Solve for D

$$D=\left(\sqrt{3}-\sqrt{2}\right)v_{0}$$