Question

I have a tricky question with 3 variables but only with 2 equations :|
\(a+b+c=45\)
\(59a+41b+40c=2008\)

P.S.: I only want hints, thank you all :)

Answer 1

a b c z ... that's 4 lol :A

Answer 2

Fixed D; It was a c, not a z >.>

Answer 3

Free variable...

Answer 4

Hint: There will not be a unique solution.

Answer 5

Trial and error? D:

Answer 6

What context is this problem in? The way I would usually solve this uses methods from linear algebra.

Answer 7

You use the same method of matrices but now you have a free variable, meaning your last variable is free to be whatever as long as it makes sense

Answer 8

Context? I'm still in 10th grade :( but I learned a bit of linear algebra.

Answer 9

By context I just mean what you're learning right now. As I'm sure you've noticed, the methods used to solve a problem tend to change a lot as you progress through learning math :)

Answer 10

Oh uh, solving equation systems with substitution, reduction and comparison, you know, basic stuffs. :)

Answer 11

You can set solution as parametric equations, like in this case,

Answer 12

Then depending on question requirement, you put in a certain value for the parameter.

Answer 13

The way we do it in linear algebra is to find the null space of the matrix that represents the system of equations, and then find a particular solution, and the solution space is the sum of a the particular solution and the null space. You can do basically the same thing without any of that terminology or matrix-oriented methodology, basically using elimination. First step, you can find a particular solution by using elimination the same way you normally would.

Answer 14

After some substitutions and eliminations
I got some interesting things
\(c=163+18a\)
\(95a + 5b=388\)
\(59a + 5b +2c= 714\)

Answer 15

And now the problem is, if I substitute the first one into \(a+b+c=45\) I would get a negative number, and from the question, a,b,x are > 0

Answer 16

Arrgh a,b,c*

Answer 17

Aha, I didn't know that the three variables had to be positive. That ends up not being a problem at all, because there is actually an infinite line of solutions. Once you have a particular solution, you can modify it to make it so that all three variables are positive.

Answer 18

Here's a hint to make this easier on you: the quickest way to find a particular solution is almost always to set one variable equal to zero, and then solve for the other two.

Answer 19

You can use parameters.

Answer 20

@nbouscal Do you mean something like stuffswithvariable = 0?

Answer 21

I just mean pick a, b, or c, and set it equal to zero. Then solve the resulting system of equations. So for example, if you set a equal to zero, you will be left with b+c=45 and 41b+40c=2008

Answer 22

Oh I see, let me try.

Answer 23

For me, I tend to write it in parameters first, then plug in suitable values for the parameters.
\[a = 9\frac{1}{18}+\frac{1}{18}t\]
\[b = 35\frac{17}{18}+1\frac{1}{18}t\]
\[c = t\]
For\[t \in \mathbb{R}\]

Answer 24

Okay, by setting up 1 variable to 0, I get
If a = 0
b+c=45
41b+40c=2008

So a=0
b=208
c=-163

Other give decimal numbers D:

Answer 25

@Kira_Yamato I didn't learn parametrization.

Answer 26

Souka... I'm sorry...

Answer 27

Yeah, I did it with a=0 also. Anyway, now you have a particular solution to the system. The next part is to learn about the null space (in linear algebra terms). Basically it provides you with a set of numbers that you can add to your solutions to get another solution. The way you find those numbers is by setting the equations equal to zero, and finding a solution. So you have a+b+c=0 and 59a+41b+40c=0. Easiest way to solve them is to use elimination. The intuition on this is actually really cool. If you find a set of numbers that, when plugged into the equations, return zero... you can add that set of numbers to a solution to the original equations and you'll still have a solution, because you're just adding zero.

Answer 28

I should point out that Kira's method of solving using parametric equations is actually a bit quicker and easier than my method if you haven't yet learned linear algebra. Once you've learned linear algebra I think my method's a little quicker, but it might be personal preference.

Answer 29

Alright now I eliminated stuffs from the equations above, I get 18b+19c=0

Answer 30

Do I have to find a set that could match the thingy above ^

Answer 31

Don't know if it helps at all
http://wiki.answers.com/Q/Is_it_possible_tofind_three_unknowns_with_two_equations

Answer 32

18b+19c=0 is correct, but isn't all of the information. You also need some information about a. What I had was that -18a+c=0, and that 19a+b=0. With that information we can find the basis of the nullspace. I had it as a=1, b=-19, and c=18. You could get it in other ways and get other numbers, but the numbers will always be multiples of those numbers. We generally represent that as \(N=c\left[
\begin{array}{c}
1\\-19\\18
\end{array}\right]\) where c is any constant. This is called a linear combination (that term makes more sense when the space in question has more than one dimension). So basically your solution space is your particular solution (a=0, b=208, c=-163) plus anything in the null space, which is any constant multiplied by the solutions above. Since we want to find a solution where all three are positive, we can set \(c=10\), so \(a=0+10\cdot1, b=208+10\cdot-19, c=-163+18\cdot10\), or \(a=10, b=18, c=17\). It turns out this is the only solution where all three variables are positive.

Answer 33

How did you find a=1, b=-19, and c=18? D:

Answer 34

Well, I had the two equations, -18a+c=0 and 19a+b=0. I got those two equations by using elimination. Once I had them, since I've been treating a as my free variable, I just set a=1 and there you go.

Answer 35

make third equation by solving these two than solve simultaneously

Answer 36

I think I understood pretty much everything here, thanks! :D

And just a little question: how to choose the constant c?

Answer 37

I just made the observation that to bring the negative solution -163 up to a positive one, we would have to have c be at least 10. 9*18=162<163.

A nice side effect of learning it this way is that when you do take more linear algebra, all of this will look kind of familiar! :)

Answer 38

Alright, you are awesome! :D Thank you very much!