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annoying simultaneous equations

c3 + 6cd2 = 99
2d3 + 3c2d = 70

Can these even be solved at all or is guesswork required?

Thanks

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Whoever thought of that question, is messing with ur head:biggrin::biggrin::biggrin:
Reply 2
Avatar for TeeEm
TeeEm
19
Original post by Ilovemaths96
c3 + 6cd2 = 99
2d3 + 3c2d = 70

Can these even be solved at all or is guesswork required?

Thanks


two ways of solving these


METHOD A

solve the fist one for d2

square the second equation and sub d2

METHOD B
use the substitution c = md or d = mc
Original post by shoriffmiah
Whoever thought of that question, is messing with ur head:biggrin::biggrin::biggrin:


Lol i was doing a step question paper 2004 and i got to that. Got the answer with guesswork lol
Original post by TeeEm
two ways of solving these


METHOD A

solve the fist one for d2

square the second equation and sub d2

METHOD B
use the substitution c = md or d = mc


What do you mean? How do you go about doing that?
Reply 5
Avatar for TirnanF
TirnanF
3
I recognise these from a step question, incredibly difficult to solve algebraically but if it is from that step question then you are told c and d are positive integers, you can see that c^3 must be less than 99, so you can narrow your possibilities down, trial and error from there until you get numbers that work for both equations.


Posted from TSR Mobile
Reply 6
Avatar for TeeEm
TeeEm
19
Original post by Ilovemaths96
What do you mean? How do you go about doing that?


you mean you cannot rearrange the first equation for d2 ?
Original post by TeeEm
you mean you cannot rearrange the first equation for d2 ?


yeah i understand what you're saying but its not gonna get me anywhere?
Original post by TirnanF
I recognise these from a step question, incredibly difficult to solve algebraically but if it is from that step question then you are told c and d are positive integers, you can see that c^3 must be less than 99, so you can narrow your possibilities down, trial and error from there until you get numbers that work for both equations.


Posted from TSR Mobile


Yeah it is a step question. In the end i guessed and my answers for c and d were correct, just thought it would be more elegant if there was an algebraic method
Reply 9
Avatar for TeeEm
TeeEm
19
Original post by Ilovemaths96
yeah i understand what you're saying but its not gonna get me anywhere?


it will definitely get me
( I do not know about you...)

I will try to find one similar question with workings from my own stuff
Original post by TeeEm
it will definitely get me
( I do not know about you...)

I will try to find one similar question with workings from my own stuff


Thanks mate, i'll try it
I like how the person suggested "guesswork". :biggrin:
Reply 12
Avatar for TeeEm
TeeEm
19
Original post by Ilovemaths96
Thanks mate, i'll try it


look at this link

http://madasmaths.com/archive_maths_booklets_standard_topics_various.html

file name

equations_advanced_skills

look at Q52 and Q53

There are some amazing equation techniques towards the end of the file
Original post by TeeEm
look at this link

http://madasmaths.com/archive_maths_booklets_standard_topics_various.html

file name

equations_advanced_skills

look at Q52 and Q53

There are some amazing equation techniques towards the end of the file


Wow, some amazing and interesting stuff on here mate, thank you very much
Reply 14
Avatar for TeeEm
TeeEm
19
Original post by Ilovemaths96
Wow, some amazing and interesting stuff on here mate, thank you very much


these questions are from my "own personal stash".
I do not normally "deal these questions"

happy times with it ...:smile:
Original post by TeeEm
these questions are from my "own personal stash".
I do not normally "deal these questions"

happy times with it ...:smile:


Haha thanks i feel so priviliged :smile:, gonna spend all christmas holidays doing these questions
Reply 16
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TeeEm
19
Original post by Ilovemaths96
Haha thanks i feel so priviliged :smile:, gonna spend all christmas holidays doing these questions


:smile:
Reply 17
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TeeEm
19
Original post by shoriffmiah
Whoever thought of that question, is messing with ur head:biggrin::biggrin::biggrin:


It is a beautiful question
(almost marriage material)
Reply 18
Avatar for TeeEm
TeeEm
19
Original post by TirnanF
I recognise these from a step question, incredibly difficult to solve algebraically but if it is from that step question then you are told c and d are positive integers, you can see that c^3 must be less than 99, so you can narrow your possibilities down, trial and error from there until you get numbers that work for both equations.


Posted from TSR Mobile


see post 13

in fact I will produce a solution 2 different ways and add this question too
Reply 19
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TeeEm
19
Original post by heavyhandscott
I like how the person suggested "guesswork". :biggrin:


:smile:

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