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Solving 2 equation with 2 variables (hard) |
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| Apr1-10, 06:00 AM | #1 |
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Solving 2 equation with 2 variables (hard)![]() I really need some help here (no program is allowed )
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| Apr1-10, 06:02 AM | #2 |
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What work have you done on the problem so far?
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| Apr1-10, 07:15 AM | #3 |
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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.
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| Apr1-10, 07:40 AM | #4 |
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Solving 2 equation with 2 variables (hard)
The problem is to find exact values, not estimatet values. Therefore: do not use any programs...
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| Apr1-10, 07:46 AM | #5 |
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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.
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| Apr1-10, 07:57 AM | #6 |
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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... |
| Apr1-10, 08:09 AM | #7 |
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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?
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| Apr1-10, 08:18 AM | #8 |
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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.
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| Apr1-10, 10:16 AM | #9 |
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There is a typo in the problem: "79" should be "78".
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| Apr1-10, 10:48 AM | #10 |
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Is it just because the answer drops our nicely if that is the case, or do you somehow know it as a fact :S ?
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| Apr1-10, 01:08 PM | #11 |
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I just somehow know it is a fact
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