Help with hard simultaneous equations

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bananaeater123
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Hey, so my friend sent me interesting simultaneous equation to solve and I can’t do it at all. I’m not sure if it’s even possible to get to them but it definitely has solutions though. It can be viewed in the picture attached. Thank you.
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Rolls_Reus_0wner
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How many marks is it big man?
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bananaeater123
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(Original post by Rolls_Reus_0wner)
How many marks is it big man?
I’m not sure, he just sent me the question and asked for help. Didn’t really say how many marks it was.
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Rolls_Reus_0wner
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(Original post by bananaeater123)
I’m not sure, he just sent me the question and asked for help. Didn’t really say how many marks it was.
I'm waiting for my friend to try solve it.

Is this UNi level?
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thekidwhogames
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(Original post by Rolls_Reus_0wner)
I'm waiting for my friend to try solve it.

Is this UNi level?
It's probably AS level or Olympiad stuff.
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Rolls_Reus_0wner
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(Original post by thekidwhogames)
It's probably AS level or Year 11 Olympiad stuff.
I do AS maths, never seen something like this.
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thekidwhogames
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(Original post by Rolls_Reus_0wner)
I do AS maths, never seen something like this.
It seems like it'd be an extremely difficult challenge or something. I've seen similar stuff on Olympiad papers.
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thekidwhogames
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Solve it like how you solve any system of equations:

Fastest method I say would be to rearrange the first equation for y:

y = (13x-x^2-21)/x^2-13

Sub this into your second equation and solve for x; plug your solutions into your rearranged expression for y.
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bananaeater123
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(Original post by Rolls_Reus_0wner)
I'm waiting for my friend to try solve it.

Is this UNi level?
(Original post by thekidwhogames)
It's probably AS level or Olympiad stuff.
Yep it’s above A-Level for sure. I’m finishing second year Further Maths and I’ve never come across one like this before.
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Rolls_Reus_0wner
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(Original post by thekidwhogames)
It seems like it'd be an extremely difficult challenge or something. I've seen similar stuff on Olympiad papers.
I did that stuff when i was like 8. Olympiad sounds familiar. But man didnt do algebra. I could barely do fractions until yr 11.
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thekidwhogames
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(Original post by bananaeater123)
Yep it’s above A-Level for sure. I’m finishing second year Further Maths and I’ve never come across one like this before.
Yeah but it may be one of those challenges or something they've got in the new spec. I wouldn't say this is university Maths but more "non-spec" stuff.
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TheOtherSide.
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Well notice that the second term for both equations is the same, so you can subtract one equation from the other to eliminate that fraction?
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bananaeater123
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(Original post by thekidwhogames)
Solve it like how you solve any system of equations:

Fastest method I say would be to rearrange the first equation for y:

y = (13x-x^2-21)/x^2-13

Sub this into your second equation and solve for x; plug your solutions into your rearranged expression for y.
Guessing you meant x^3 on the numerator and it’s not that easy. You substitute in to the second and then you have a polynomial with a degree of 6. I don’t know how you solve those if there’s even a method that’s not numerical or factorisation. That’s why I thought maybe there was an easier way.
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bananaeater123
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(Original post by TheOtherSide.)
Well notice that the second term for both equations is the same, so you can subtract one equation from the other to eliminate that fraction?
Yep then you can get an expression for x or y but when you substitute it back in you have square roots of x or y everywhere that don’t cancel or anything.
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Rolls_Reus_0wner
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nearly there mate
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thekidwhogames
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(Original post by bananaeater123)
Guessing you meant x^3 on the numerator and it’s not that easy. You substitute in to the second and then you have a polynomial with a degree of 6. I don’t know how you solve those if there’s even a method that’s not numerical or factorisation. That’s why I thought maybe there was an easier way.
Yeah sorry I made a typo. Apart from numerical methods, I haven't see another method. I'll try it again to find an easier solution.
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AmmarTa
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You can get an equation for x or y and substitute into another (I'll try it after tea).
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RDKGames
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No point attempting these by hand. Numerical methods is the way.
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Rolls_Reus_0wner
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Man there r no solutions
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DFranklin
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(Original post by RDKGames)
No point attempting these by hand. Numerical methods is the way.
It wasn't obvious to me this can't be done by hand, but numerically I find:

x = 3.1433621488
y = 3.5889727776

and the inverse symbolic calculator can't find anything analytic that matches, so yeah, doesn't look like this is doable by hand...
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