
#1
Apr110, 06:00 AM

P: 35

I really need some help here (no program is allowed ) 



#3
Apr110, 07:15 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

I'm not sure why you would consider that "hard". You can solve the second equation for x as a quadratic function of y. Putting that into the first equation gives a cubic equation for y.




#4
Apr110, 07:40 AM

P: 35

Solving 2 equation with 2 variables (hard)
The problem is to find exact values, not estimatet values. Therefore: do not use any programs...




#5
Apr110, 07:46 AM

HW Helper
P: 3,353

Indeed. Do you have any strategies to find a root? There are many you can try. Once you find one, you can divide and just solve a quadratic.




#6
Apr110, 07:57 AM

P: 35

Here is what I have done so far:
But these solutions contains "cos". It is possible to rewrite the equations to the formula and then find solutions without any trigonometrical components. That is my problem... 



#7
Apr110, 08:09 AM

HW Helper
P: 3,353

Would have been a simpler computation for the program to have changed the x into 1/y rather than the other way around, small numbers to deal with overall. Were you expected to solve this cubic analytically and it's roots actually looked like that?




#8
Apr110, 08:18 AM

P: 35

Yes. It does not matter how you start. You will get the same solutions with the same trigonometrical components. Plz help me find solutions without any trigonometrical components.




#9
Apr110, 10:16 AM

P: 2,159

There is a typo in the problem: "79" should be "78".




#10
Apr110, 10:48 AM

HW Helper
P: 3,353

Is it just because the answer drops our nicely if that is the case, or do you somehow know it as a fact :S ?




#11
Apr110, 01:08 PM

P: 35

I just somehow know it is a fact



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