Common denominator with radicals

Question

Can you make these 2 fractions into 1?

$$2\sqrt{9-2x} - \dfrac{2x}{\sqrt{9-2x}}$$

I thought you could make them into $ \dfrac{-2x+18}{\sqrt{9-2x}}$

Answer 1

You're almost correct.

\begin{align} 2\sqrt{9-2x} - \frac{2x}{\sqrt{9-2x}} &= 2\sqrt{9-2x} \times \frac{\sqrt{9-2x}}{\sqrt{9-2x}} - \frac{2x}{\sqrt{9-2x}}\\ &= \frac{2(\sqrt{9-2x})^2}{\sqrt{9-2x}} - \frac{2x}{\sqrt{9-2x}}\\ &= \frac{2(9-2x)}{\sqrt{9-2x}} - \frac{2x}{\sqrt{9-2x}}\\ &= \frac{2(9-2x) - 2x}{\sqrt{9-2x}}\\ &= \frac{18 - 4x - 2x}{\sqrt{9-2x}}\\ &= \frac{18-6x}{\sqrt{9-2x}} \end{align}