Cannot figure out how to solve for x, y: sqrt{x} - sqrt{2y} = sqrt{12} - sqrt{8}; 4^{x/y} - 3*4^{(5y-x)/y} = 16

[gamaz321]

I got a simultaneous equation that is rather complex. I could do just a few steps of the problem. After these steps, I have no idea how to proceed. Any idea or help is highly appreciated. Thank you in advance.

 

[Steven G]

Which is it, the 2y is under the radical or is just the 2 under the radical? You have written it both ways!

Step 1: do the long division for the exponent of 4.

 

[gamaz321]

2y is under the radical. Sorry for the confusion.

 

[Steven G]

2y is under the radical. Sorry for the confusion.

OK, but since one line had sqrt(2y) and the next line has it as sqrt(2)*y there must be a problem with what you wrote.

 

[BigBeachBanana]

Make a substitution [imath]u = \dfrac{x}{y}[/imath]
[math]\tag{1}4^u+3\times4^{5-u}=16[/math][math]\tag{2}\sqrt{u}+\sqrt{2}= \dfrac{\sqrt{12}-\sqrt{8}}{\sqrt{y}}[/math]
You can solve for [imath]u[/imath] from equation (1) and use equation (2) to solve for [imath]y[/imath], then [imath]x[/imath] follows.

 

[gamaz321]

Make a substitution [imath]u = \dfrac{x}{y}[/imath]
[math]\tag{1}4^u+3\times4^{5-u}=16[/math][math]\tag{2}\sqrt{u}+\sqrt{2}= \dfrac{\sqrt{12}-\sqrt{8}}{\sqrt{y}}[/math]
You can solve for [imath]u[/imath] from equation (1) and use equation (2) to solve for [imath]y[/imath], then [imath]x[/imath] follows.

Thank you for your help! It worked well! Thank you.

 

[khansaheb]

Thank you for your help! It worked well! Thank you.

Please share your solution - so that it can teach other students.

 

[gamaz321]

Sure. Here is the solution in the attachment. Hope it helps those who would find interest in this tricky problem.

 

[gamaz321]