| | Trig help!!!! |  |
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Ahmed
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 |  Subject: Trig help!!!!  Wed Sep 09, 2009 6:09 pm | |
| PLEASE HELP! I am going crazy.
If x+y=1, x^2+y^2=4, x^3+y^3=?
^x=exponent
Lol, first day of Advanced Algebra and Trig and I feel like I'm lost -___-
Please explain slowly. | |
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Fangy
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| | Subject: Re: Trig help!!!! Wed Sep 09, 2009 6:15 pm | |
| Lol, makes no sense to me. If you take the square roots of the second one, it would mean that x+y=2... | |
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Ahmed
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| | Subject: Re: Trig help!!!! Wed Sep 09, 2009 6:18 pm | |
| This isn't simple Algebra that's why it doesn't sense to you (or me). | |
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preid1220
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| | Subject: Re: Trig help!!!! Wed Sep 09, 2009 7:13 pm | |
| well if it follows a set pattern then the answer may be 6, it could also be 9. seeing as i failed trig it may not be that useful to listen to me tho | |
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Fangy
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| | Subject: Re: Trig help!!!! Wed Sep 09, 2009 7:55 pm | |
| Is that actually trig? I think that's just algebra... I don't see anything related to trig in it (I myself am taking trig now) | |
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preid1220
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| | Subject: Re: Trig help!!!! Wed Sep 09, 2009 8:04 pm | |
| - Fang wrote:
- Is that actually trig? I think that's just algebra... I don't see anything related to trig in it (I myself am taking trig now)
trig and algebra go hand in hand | |
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Ahmed
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| | Subject: Re: Trig help!!!! Wed Sep 09, 2009 9:48 pm | |
| Well it's "Advanced Algebra and Trig", this is also known as "Pre-Calculus" | |
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Fangy
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| | Subject: Re: Trig help!!!! Thu Sep 10, 2009 12:14 am | |
| Not trig then.
I'll post it on Sal's and see if they can help. | |
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Fangy
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| | Subject: Re: Trig help!!!! Thu Sep 10, 2009 4:23 pm | |
| Here it is:
How come you changed the result of the first equation into 2 rather than 1 ?
I'd substitute for x or y from the first equation:
x+y=1 --> y=1-x
Then substitute y into the second equation and solve for x as a quadratic. Does that help? I get two possible values for each of x and y, but I'll leave you to work them out for yourself... | |
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Ahmed
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| | Subject: Re: Trig help!!!! Thu Sep 10, 2009 7:06 pm | |
| - Fang wrote:
- Here it is:
How come you changed the result of the first equation into 2 rather than 1 ?
I'd substitute for x or y from the first equation:
x+y=1 --> y=1-x
Then substitute y into the second equation and solve for x as a quadratic. Does that help? I get two possible values for each of x and y, but I'll leave you to work them out for yourself... No sorry. Thanks though  I finally understood it today  If x+y=1, x^2+y^2=4, x^3+y^3=? x^3+y^3=(x+y)(x^2-xy+y^2) x^3+y^3=(1)(4-xy) ---------- (x+y)^2=x^2+2xy+y^2 x^2+2xy+y^2=1 (4)+2xy=1 2xy=-3 xy=-3/2 --------- x^3+y^3=(1)(4-xy) x^3+y^3=(1)(4-(-3/2)) x^3+y^3=(1)(4+3/2) x^3+y^3=(1)(11/2) x^3+y^3=11/2 | |
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Fangy
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| | Subject: Re: Trig help!!!! Thu Sep 10, 2009 9:24 pm | |
| Just btw that was from Slashing UK | |
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Ahmed
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| | Subject: Re: Trig help!!!! Thu Sep 10, 2009 9:33 pm | |
| He probably saw it as a normal Algebra question... | |
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