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# C1 Equations

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1. Hi!
I am not sure how to do the first part of this question. I have tried to eliminate x but I did not get the right answer.

Thanks
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2. Try to rearrange equation (1) to find a variable equated to something. Then substitute that variable's rearrangement into equation (2).
3. (Original post by musicangel)
Hi!
I am not sure how to do the first part of this question. I have tried to eliminate x but I did not get the right answer.

Thanks
Multiply the first equation by 3y then subtract the two. Should be straight forward.
Try to rearrange equation (1) to find a variable equated to something. Then substitute that variable's rearrangement into equation (2).
Mate it's explicitly asking to do this via elimination rather than substitution.
5. (Original post by RDKGames)
Mate it's explicitly asking to do this via elimination rather than substitution.
I used to think elimination is crossing numbers out then i got moved to set 2 maths in gcse

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6. (Original post by RDKGames)
Mate it's explicitly asking to do this via elimination rather than substitution.
Substitution is a form of elimination, I don't see why it wouldn't be acceptable here. You're still eliminating .
7. (Original post by Zacken)
Substitution is a form of elimination, I don't see why it wouldn't be acceptable here. You're still eliminating .
Beefting

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8. (Original post by Zacken)
Substitution is a form of elimination, I don't see why it wouldn't be acceptable here. You're still eliminating .
True but you know how those C1 books are. Either you do it your way, or their way. I'm assuming they want it done by adding the two equations as it would follow straight from GCSE's methods quite well. Personally I wouldn't care which way as long as the solutions are obtained.
9. (Original post by musicangel)
Hi!
I am not sure how to do the first part of this question. I have tried to eliminate x but I did not get the right answer.

Thanks
Could you post your working so we can see where you have gone wrong?
10. Thank you all for your help - I now know my mistake.
With this question, I am not sure where I have gone wrong so would really appreciate any help

Thanks
Attached Images

11. (Original post by musicangel)
Thank you all for your help - I now know my mistake.
With this question, I am not sure where I have gone wrong so would really appreciate any help

Thanks
However you haven't factorised it properly, as if you multiply out what you put, you get 8y^2-9y-14, so the 9 and 14 are the wrong way round.
The correct factorisation is (2y+1)(4y-9).
12. (Original post by RDKGames)
Multiply the first equation by 3y then subtract the two. Should be straight forward.
I know the lines are blurred but I think you've given a bit too much information here, when you compare it to the post above at least. I'm sorry if this feels like I'm singling you out - I'm not - but this follows from discussions on previous threads. And again, very much appreciate your enthusiasm/efforts - it's something I had to get used to doing when I first started.

I would have asked them to show their working for how they tried to eliminate x before suggesting anything.
13. (Original post by SeanFM)
I know the lines are blurred but I think you've given a bit too much information here, when you compare it to the post above at least. I'm sorry if this feels like I'm singling you out - I'm not - but this follows from discussions on previous threads. And again, very much appreciate your enthusiasm/efforts - it's something I had to get used to doing when I first started.
Perhaps, I understand. As you said, lines are blurred, and I do think my information was as much as the post's above rather than more, just a shorter method.
14. (Original post by HapaxOromenon3)
However you haven't factorised it properly, as if you multiply out what you put, you get 8y^2-9y-14, so the 9 and 14 are the wrong way round.
The correct factorisation is (2y+1)(4y-9).
Thank you!
With the question attached (4b), the answers only say p ≤ 8. Why does it not include p≥0? Also, why is it 'less than and equal to' instead of just less than?
Thanks
Attached Images

15. (Original post by musicangel)
Thank you!
With the question attached (4b), the answers only say p ≤ 8. Why does it not include p≥0? Also, why is it 'less than and equal to' instead of just less than?
Thanks
Because a quadratic gives 2 distinct real roots if the discriminant (b2-4ac) is more than 0. It gives equal REAL roots if it is equal to 0, and two imaginary roots if it is less than 0. As you can see, you can get real roots if the discriminant is either 0 or more than 0, hence the inequality.

And the answer should also include p>0 as well. Can't be equal to 0 otherwise it's not a quadratic.
16. (Original post by RDKGames)
And the answer should also include p>0 as well. Can't be equal to 0 otherwise it's not a quadratic.

That's not the reason, nobody said it was a quadratic in the first place. The real reason why is because .
17. (Original post by Zacken)
That's not the reason, nobody said it was a quadratic in the first place. The real reason why is because .
Well I'm sure we have to assume it is a quadratic in order to apply the discriminant of the quadratic then narrow the range for p further by inspection.

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18. (Original post by RDKGames)
Well I'm sure we have to assume it is a quadratic in order to apply the discriminant of the quadratic then narrow the range for p further by inspection.

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Exactly why what I said in my post stands.
19. (Original post by Zacken)
Exactly why what I said in my post stands.
Sure, but I'm saying that if p=0 then we don't have a quadratic which contradicts the original assumption. Of course, -2=0 is also another way to explain the nonsense. So I don't see why it wouldn't be a reason.
20. (Original post by RDKGames)
Sure, but I'm saying that if p=0 then we don't have a quadratic which contradicts the original assumption. Of course, -2=0 is also another way to explain the nonsense. So I don't see why it wouldn't be a reason.
Yes, but your assumption is unjustified. Basically, you're unconciously splitting into cases:

Case (i): assume p =/= 0 then quadratic and discriminant

Case (ii): assume p = 0 then ....

You can't just randomly assume p =/= 0 because you feel like and then go well "p =0 contradicts the original assumption", of course it does, you assumed p =/= 0 for no reason at all, so obviously p=0 would contradict that. You need to go back to the root and deal with the p = 0 case on its own.

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