Help! working order algebra

[seppege]

Hey guys, I'm stuck at this last question of my online math test, i'm free to use help. You have to write it as simple as possible and one of the following 4 awnsers is correct. I've been struggling with this for a while now and can't seem to figure out the right working order. I tried putting all den. to 6 first then solve the the x and y's but I can't get the square root 6 away on the bottom, or my awnser is nowhere near any of these 4 . Any help with working order appreciated! thanks

 

[HallsofIvy]

Well, first, all of the "x"s and "y"s are to the same power in all answers so that part can be ignored. You say "I can't get the square root 6 away". You know, of course, that 6= 2(3) so \(\displaystyle \sqrt{6}= \sqrt{2}\sqrt{3}\) and you can cancel part of the "square root 6".

The first coefficient is \(\displaystyle \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{6}}{3}}= \frac{\sqrt{3}}{2}\frac{3}{\sqrt{3}\sqrt{2}}= \frac{3}{2\sqrt{2}}\) because the two "\sqrt{3}"s cancel. Now, "rationalizing the denominator" by multiplying both numerator and denominator by \(\displaystyle \sqrt{2}\) gives \(\displaystyle \frac{3\sqrt{2}}{2(2)}= \frac{3\sqrt{2}}{4}\). That is the fourth option.

 

[seppege]

Well, first, all of the "x"s and "y"s are to the same power in all answers so that part can be ignored. You say "I can't get the square root 6 away". You know, of course, that 6= 2(3) so \(\displaystyle \sqrt{6}= \sqrt{2}\sqrt{3}\) and you can cancel part of the "square root 6".

The first coefficient is \(\displaystyle \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{6}}{3}}= \frac{\sqrt{3}}{2}\frac{3}{\sqrt{3}\sqrt{2}}= \frac{3}{2\sqrt{2}}\) because the two "\sqrt{3}"s cancel. Now, "rationalizing the denominator" by multiplying both numerator and denominator by \(\displaystyle \sqrt{2}\) gives \(\displaystyle \frac{3\sqrt{2}}{2(2)}= \frac{3\sqrt{2}}{4}\). That is the fourth option.

Hi thanks for your quick reply, I understand how it works now. Just to make sure it's never a good idea to put all denominators to 6 first? With your help for the final exercise I got a 10/10 cheers!!!