Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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}}$

share|improve this question
rather $18-4x-2x$ in the nominator –  Berci Oct 13 '12 at 11:35
You're right, thank you. –  Tittyboy Oct 13 '12 at 11:38
Else it's maximally ok –  Berci Oct 13 '12 at 11:39
add comment

1 Answer

up vote 2 down vote accepted

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}

share|improve this answer
I assume that we should make the statement that $x$ is not equal 9/2. –  Emmad Kareem Oct 13 '12 at 12:38
That is implicit as soon as the initial expression is given. In fact, you need $x < \frac{9}{2}$. –  Michael Albanese Oct 13 '12 at 12:44
OK, thanks, may be the OP should take notice. –  Emmad Kareem Oct 13 '12 at 12:45
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.