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Finding three unknowns from two equations

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No, it was me admiring your patience. I haven't worked out how to attach a message to accompany the +1 (or are they hoping to sell more subs?).

You can't attach messages to post +1s, but you can positive rep people on their profiles, which is like the old rep system, and you can attach comments. You can do one profile rep (which can only be positive) and five post reps (either pos or neg) per day, unless you're a sub, who can do one profile rep and ten post reps.

On topic: I hope you guys see now that this is not a solveable set of equations. I find that a good way to deal with people who say "that is easy to solve!" is to say "ok then, can you solve it for me?" ... they never post back after that (see Carter_C and Peace0fM1nd).

Reply 41

dknt

I think we should probably Latex-ify the equations to stop further confusion...

\frac{9}{5}a + \frac{7}{5}b +\frac{4}{5}c = 0

-2a +\frac{3}{5}b +\frac{4}{5}c = 0

This is as far as I understand. If not, it'll have been a futile effort.

Reply 42

Mr M

Study Forum Helper

Original post by dknt

I think we should probably Latex-ify the equations to stop further confusion...

\frac{9}{5}a + \frac{7}{5}b +\frac{4}{5}c = 0

-2a +\frac{3}{5}b +\frac{4}{5}c = 0

This is as far as I understand. If not, it'll have been a futile effort.

This is like strapping an oxygen mask to a patient who died last week.

Reply 43

Peace0fM1nd

Original post by M_E_X

err.... im nearly ther with a solution using another method. patience

Reply 44

M_E_X

Original post by dknt

I think we should probably Latex-ify the equations to stop further confusion...

\frac{9}{5}a + \frac{7}{5}b +\frac{4}{5}c = 0

-2a +\frac{3}{5}b +\frac{4}{5}c = 0

This is as far as I understand. If not, it'll have been a futile effort.

That's actually useful, thank you :smile:

I think those equations are right too - some people thought the unknowns were on the bottom, but most people agree they are likely to be on the top.

This should help clarify that, thank you :smile:

Reply 45

dknt

Original post by M_E_X

That's actually useful, thank you :smile:

I think those equations are right too - some people thought the unknowns were on the bottom, but most people agree they are likely to be on the top.

This should help clarify that, thank you :smile:

No problem :smile:

Original post by Peace0fM1nd

err.... im nearly ther with a solution using another method. patience

I'd quite like to see where you are with it. I sincerely hope you're not trying a trial and improvement approach.

Reply 46

Pheylan

i think that we should prepare for the possibility of the questioner giving us a third equation. let this equation be

pa+qb+rc=s

, then we have:

9a+7b+4c=0

-10a+3b+4c=0

pa+qb+rc=s

\begin{pmatrix} 9 & 7 & 4 \\-10 & 3 & 4 \\p & q & r \end{pmatrix}\begin{pmatrix} a \\b \\c \end{pmatrix}=\begin{pmatrix} 0 \\0 \\s \end{pmatrix}

with a little bit of calculation

\begin{pmatrix} a \\b \\c \end{pmatrix}=\dfrac{1}{16p-76q+97r}\begin{pmatrix} 3r-4q & 4q-7r & 16 \\4p+10r & 9r-4p & -76 \\-3p-10q & 7p-9q & 97 \end{pmatrix}\begin{pmatrix} 0 \\0 \\s \end{pmatrix}

Reply 47

M_E_X

Original post by Pheylan

i think that we should prepare for the possibility of the questioner giving us a third equation. let this equation be

pa+qb+rc=s

, then we have:

9a+7b+4c=0

-10a+3b+4c=0

pa+qb+rc=s

\begin{pmatrix} 9 & 7 & 4 \\-10 & 3 & 4 \\p & q & r \end{pmatrix}\begin{pmatrix} a \\b \\c \end{pmatrix}=\begin{pmatrix} 0 \\0 \\s \end{pmatrix}

with a little bit of calculation

\begin{pmatrix} a \\b \\c \end{pmatrix}=\dfrac{1}{16p-76q+97r}\begin{pmatrix} 3r-4q & 4q-7r & 16 \\4p+10r & 9r-4p & -76 \\-3p-10q & 7p-9q & 97 \end{pmatrix}\begin{pmatrix} 0 \\0 \\s \end{pmatrix}

Great - once you've got some answers, let me know!

Remember that we know we can get an answer in terms of the other two coefficients, but we don't think you can get actual values for your answers (well, not unique ones anyway - there are an infinite number of them, after all...)

Reply 48

Mr M

Study Forum Helper

Original post by M_E_X

there are an infinite number of them, after all...)

This thread is steadfastly refusing to die. I think we should also prepare for the possibility that some intrepid soul may attempt to list every solution.

Reply 49

M_E_X

Original post by Mr M

This thread is steadfastly refusing to die. I think we should also prepare for the possibility that some intrepid soul may attempt to list every solution.

That is becoming a very real possibility. I hope that the new vB 3.8 upgrade can handle posts of an infinite length.

Reply 50

Pheylan

i wasn't being serious. i should've made that clearer.

Reply 51

Dirac Spinor

Original post by Pheylan

i wasn't being serious. i should've made that clearer.

We were all too in awe of your maths/Latex ninjery

Reply 52

M_E_X

Original post by Peace0fM1nd

err.... im nearly ther with a solution using another method. patience

Did you get a solution? It's been over ten hours since you asked for patience, I hope you managed to get some sleep!

Reply 53

Peace0fM1nd

Original post by M_E_X

Did you get a solution? It's been over ten hours since you asked for patience, I hope you managed to get some sleep!

lmao - it sure is a difficult 1. i did get some sleep in the end and apologies for not letting u know. hav u made any progress yet?? i got close to a solution last nite but hit a brick wall right in the end.

p.s. i guess we'l have to ask wolfram alpha

Reply 54

Peace0fM1nd

Wolfram Alpha solution:

http://www.wolframalpha.com/input/?i=%289%2F5%29a%2B%287%2F5%29b%2B%284%2F5%29c%3D%28-2%29a%2B%283%2F5%29b%2B%284%2F5%29c

Reply 55

M_E_X

Original post by Peace0fM1nd

It is unsolvable - do you still not agree? We have three equations and two unknowns.

Reply 56

Peace0fM1nd

Original post by M_E_X

It is unsolvable - do you still not agree? We have three equations and two unknowns.

Its clearly solvable from wolfram alpha and i think we all agree it would have took a genius to solve it unaided... also why are you solving this anyway is it for uni?

Reply 57

M_E_X

Original post by Peace0fM1nd

Its clearly solvable from wolfram alpha and i think we all agree it would have took a genius to solve it unaided... also why are you solving this anyway is it for uni?

Your wolfram alpha only provided two values, for a and b. There are an infinite number of solutions to this equation - we cannot find a unique solution, so we say it is not solvable.

Let me try to explain it to you again:

If we had the equation

a - 5 = 10, using Wolfram alpha or Gaussian elimination or however you want to solve that,we can find a = 15.

Now if we have the equation

a + b = 20 - how can we solve that? There are an infinite number of solutions, seeing as we can pick any value for a and choose a value for b to make the equation hold. It's a similar situation with the set of equations in the OP: there aren't enough equations to 'pin down' values of all the unknowns, so the problem is unsolvable.

And it's not uni-level, it's probably GCSE or at best A-level, probably GCSE as a test of 'do you know you need n equations to solve for n unknowns' ruleb(which a lot of people here, including you, apparently don't).

Reply 58

Farhan.Hanif93

Original post by Peace0fM1nd

Its clearly solvable from wolfram alpha and i think we all agree it would have took a genius to solve it unaided... also why are you solving this anyway is it for uni?

Has it given numerical values for a,b and c? NO. So there is no unique solution.
No matter how smart we become, without completely changing the laws of maths, we will never be able to give a numerical solution for a,b and c.
Wolfram has NOT done this so it hasn't solved our problem.