Scyzo wrote:

Thanks! But could you elaborate on the last part a bit more? I do not quite get how you isolated x or y in the end...

Hi Scyzo",

Sure. There was an error in my last post. The value of x is 16/3 and y is 1/4. xy is still equal to 4/3. Sorry for that.

We have following equations:

\(\sqrt{3x}-\sqrt{4y}=3\) --- (1)

and

\(\sqrt{3x}+\sqrt{4y} = 5\) --- (2)

Add equation (1) and (2) we have following

\(2 \sqrt{3x} = 8 \Rightarrow \sqrt{3x} = 4\). By squaring both side, we have \(3x = 16 \Rightarrow x = 16/3\)

Now substitute the value of x in equation (2).

\(\sqrt{3 \times \frac{16}{3}} + \sqrt{4y} = 5 \Rightarrow \sqrt{4y} = 5 - 4 = 1\)

By squaring both sides, we have \(4y = 1 \Rightarrow y = \frac{1}{4}\).

\(xy = \frac{16}{3} \times \frac{1}{4} = \frac{4}{3}\)

Hope this helps. Please let me know, if something is not clear.