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 May 26, 2014

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 #1
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\frac{c}{\sqrt{2x}}+\frac{y}{2\sqrt{x}}Multiplytopandbottomof1sttermby$2$\frac{c}{\sqrt{2x}}+\frac{y}{2\sqrt{x}}=\frac{\sqrt2c}{\sqrt2\sqrt{2x}}+\frac{y}{2\sqrt{x}}Thedenominatorofthefirsttermcanberewrittenas$22x=2x$so \frac{c}{\sqrt{2x}}+\frac{y}{2\sqrt{x}}=\frac{\sqrt2c}{2\sqrt{x}}+\frac{y}{2\sqrt{x}}Bothtermsnowhaveacommondenominatorso:\frac{c}{\sqrt{2x}}+\frac{y}{2\sqrt{x}}=\frac{\sqrt2c+y}{2\sqrt{x}}$$

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 May 26, 2014
 #1
avatar+33654 
+5
Best Answer

\frac{c}{\sqrt{2x}}+\frac{y}{2\sqrt{x}}Multiplytopandbottomof1sttermby$2$\frac{c}{\sqrt{2x}}+\frac{y}{2\sqrt{x}}=\frac{\sqrt2c}{\sqrt2\sqrt{2x}}+\frac{y}{2\sqrt{x}}Thedenominatorofthefirsttermcanberewrittenas$22x=2x$so \frac{c}{\sqrt{2x}}+\frac{y}{2\sqrt{x}}=\frac{\sqrt2c}{2\sqrt{x}}+\frac{y}{2\sqrt{x}}Bothtermsnowhaveacommondenominatorso:\frac{c}{\sqrt{2x}}+\frac{y}{2\sqrt{x}}=\frac{\sqrt2c+y}{2\sqrt{x}}$$

Alan May 26, 2014
 #2
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thank you very much :)

 May 26, 2014

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