### Instruction: Rationalize the denominator of $\frac{\sqrt{8}+\sqrt{3}}{\sqrt{2}+\sqrt{3}}$. Express your answer in simplest form.### Response: Rationalizing the denominator means getting rid of any square roots in the bottom of the fraction.
A common technique for doing this is to multiply the fraction by a conjugate of the denominator, which is the same expression but with the sign changed.
This way, we can use the difference of squares formula to simplify the product.
For example, $(a+b)(a-b) = a^2 - b^2$.
So, the conjugate of $\sqrt{2}+\sqrt{3}$ is $\sqrt{2}-\sqrt{3}$.
If we multiply the fraction by $\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}$, which is just 1, we don't change its value, but we do rationalize the denominator.
Let's see what happens.
We get $\frac{\sqrt{8}+\sqrt{3}}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} = \frac{(\sqrt{8}+\sqrt{3})(\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}$.
Now, we need to multiply out the numerators and denominators.
For the numerator, we can use the distributive property.
We get $(\sqrt{8}+\sqrt{3})(\sqrt{2}-\sqrt{3}) = \sqrt{8}\sqrt{2} - \sqrt{8}\sqrt{3} + \sqrt{3}\sqrt{2} - \sqrt{3}\sqrt{3}$.
We can simplify some of these terms by using the property of radicals that $\sqrt{ab} = \sqrt{a}\sqrt{b}$.
For example, $\sqrt{8}\sqrt{2} = \sqrt{8 \times 2} = \sqrt{16} = 4$.
Similarly, $\sqrt{3}\sqrt{3} = \sqrt{9} = 3$.
So, the numerator becomes $4 - \sqrt{24} + \sqrt{6} - 3$.
We can combine the constant terms and get $1 - \sqrt{24} + \sqrt{6}$.
We can also simplify $\sqrt{24}$ by factoring out a perfect square.
We get $\sqrt{24} = \sqrt{4 \times 6} = \sqrt{4} \sqrt{6} = 2\sqrt{6}$.
So, the numerator becomes $1 - 2\sqrt{6} + \sqrt{6} = 1 - \sqrt{6}$.
For the denominator, we can use the difference of squares formula.
We get $(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3}) = \sqrt{2}^2 - \sqrt{3}^2 = 2 - 3 = -1$.
So, the fraction becomes $\frac{1 - \sqrt{6}}{-1}$.
We can simplify this by dividing both the numerator and denominator by -1.
We get $\frac{-1 + \sqrt{6}}{1} = -1 + \sqrt{6}$.