How do you do this?
We have 9/5a + 7/5b +4/5c = 0
and 2a +3/5b +4/5c = 0
I know it can be done, but I can't seem to figure it out!
Finding three unknowns from two equations
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How do you do this?
We have 9/5a + 7/5b +4/5c = 0
and 2a +3/5b +4/5c = 0
I know it can be done, but I can't seem to figure it out! 
You posted this yesterday and were advised you need a third equation to find a unique solution.

(Original post by TimothyTankT)
How do you do this?
We have 9/5a + 7/5b +4/5c = 0
and 2a +3/5b +4/5c = 0
I know it can be done, but I can't seem to figure it out!
You need one equation for each unknown, so as Mr M said you'd need another equation here. 
I am sure I saw somewhere that you can use some sort of iterative method or something. It won't get you the exact answer, but it'll give you an aproximation.

Definitely you need a third equation.

(Original post by StephenP91)
I am sure I saw somewhere that you can use some sort of iterative method or something. It won't get you the exact answer, but it'll give you an aproximation. 
(Original post by M_E_X)
There are an infinite number of solutions of a, b and c that will solve both of those equations. To pin down one solution you need a third equation  really really.
What's this all about then? 
(Original post by M_E_X)
There are an infinite number of solutions of a, b and c that will solve both of those equations. To pin down one solution you need a third equation  really really. 
(Original post by StephenP91)
http://answers.yahoo.com/question/in...5210425AAPWAaH
What's this all about then?
Let me put it in to context for you:
If we had the equation
a + b = 10 , how many solutions does that have? An infinite number, we can choose any value of a and choose a value of b accordingly.
Similarly, in our example, we have too many free variables (3 unknowns and only two equations), and you could rearrange it in to the situation above where we have two unknowns and only one equation. To do this you could eliminate c by rearranging the bottom equation to c=... and then subbing that for c in to the top equation. We'd then have one equation with two unknowns  unsolvable. 
(Original post by Mr M)
Chuck Norris could do it with two. 
(Original post by M_E_X)
He has approximated one of the (infinite) number of solutions.
Let me put it in to context for you:
If we had the equation
a + b = 10 , how many solutions does that have? An infinite number, we can choose any value of a and choose a value of b accordingly.
Similarly, in our example, we have too many free variables (3 unknowns and only two equations), and you could rearrange it in to the situation above where we have two unknowns and only one equation. To do this you could eliminate c by rearranging the bottom equation to c=... and then subbing that for c in to the top equation. We'd then have one equation with two unknowns  unsolvable. 
(Original post by StephenP91)
I know, but what's stopping using one of those answers?
As I wrote in my first post, it's easy to solve this for a and b in terms of c, but it's impossible to find a solution for all three variables.
If we had the equation a + b = 10, and I said a=7 and b=3, that is an answer but it's not satisfactory because there are an infinite number of other solutions. It would be better to just express it as a = 10  b and b = 10a, rather than picking one solution. 
Surely you don't need a 3rd equation for that?
They both =0 which means you can equate them:
9/5a + 7/5b +4/5c = 2a +3/5b +4/5c
Subtract 4/5c from both sides:
9/5a + 7/5b = 2a + 3/5b
And with some messing about you could eventually work out the unknowns? 
(Original post by Carter_C)
Surely you don't need a 3rd equation for that?
They both =0 which means you can equate them:
9/5a + 7/5b +4/5c = 2a +3/5b +4/5c
Subtract 4/5c from both sides:
9/5a + 7/5b = 2a + 3/5b
so:
19/5a = 4/5b
19a = 4b
And with some messing about you could eventually work out the unknowns? 
Actually thinking about it a bit more, I'm a complete dick
..but you can work out that a =/= 0
So in a way I'm right 
(Original post by Carter_C)
Actually thinking about it a bit more, I'm a complete dick
..but you can work out that a =/= 0
So in a way I'm right
As I said, there are an infinite number of solutions, and a=0 in some of them. 
(Original post by M_E_X)
Sometimes a=0.
As I said, there are an infinite number of solutions, and a=0 in some of them. 
(Original post by Carter_C)
If a is zero the first equation won't work babez
I'll work it out for you seeing as you didn't seem to get it last time.
If a=0, b=0 and c=0
9/5a + 7/5b +4/5c = 0
Subbing in (that means replacing a, b and c all by zero)
9/5(0) + 7/5(0) +4/5(0) = 0
0 + 0 + 0 = 0
0 = 0
The equation holds, therefore a=0, b=0 and c=0 is a valid solution.
"babez". 
i just did some long algebra. i did use the calc a bit though (to speed things up), idk if you can.
im a noob at maths though, so i'll just say i got:
b = 1
a = 4/19
c = 17/76
pretty sure i got it wrong though.
so if you know the answers and see this wrong, do NOT do what i did which was:
find a in terms of b.
use a in terms of b to find b.
sub b in 'a in terms of b' to find a.
sub these into one equation.
yeah. sorry if that was long.
*Edit* mega lag whilst trying to edit >__> yeah i checked this and my answers are WRONG. they only work for the second equation and not the first ¬¬ sorry OP. 
(Original post by M_E_X)
If a=0, b=0 and c=0, then what?
I'll work it out for you seeing as you didn't seem to get it last time.
If a=0, b=0 and c=0
9/5a + 7/5b +4/5c = 0
Subbing in (that means replacing a, b and c all by zero)
9/5(0) + 7/5(0) +4/5(0) = 0
0 + 0 + 0 = 0
0 = 0
The equation holds, therefore a=0, b=0 and c=0 is a valid solution.
"babez".
I think you're interpreting 9/5a as being (9/5)a , which is retarded on your part.
Think again babez
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