C1 Equations

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.

(edited 7 years ago)

Reply 3

Original post by ETbuymilkandeggs

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.

Reply 4

username1162972

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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Reply 5

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

x

Reply 6

username1162972

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

x

Beefting

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Reply 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

x

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.

(edited 7 years ago)

Reply 8

Muttley79

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?

Reply 9

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

(edited 7 years ago)

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

Your quadratic is correct: 8y^2-14y-9=0.
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).

Reply 11

Kevin De Bruyne

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.

(edited 7 years ago)

Reply 12

Original post by SeanFM

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.

Reply 13

musicangel

Original post by HapaxOromenon3

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

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.

Reply 15

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

p \neq 0

is because

-2 \neq 0

Reply 16

Original post by Zacken

That's not the reason, nobody said it was a quadratic in the first place. The real reason why

p \neq 0

is because

-2 \neq 0

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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(edited 7 years ago)

Reply 17

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.

Reply 18

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.

Reply 19