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Question about linear equation

by icesalmon
Tags: equation, linear
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icesalmon
#1
Nov27-12, 04:43 PM
P: 173
1. The problem statement, all variables and given/known data
Show that if the linear equations x1 + kx2 = c and x1 + lx2 = d have the same solution set, then the equations are identical ( k = l, c = d)

3. The attempt at a solution
since the two equations have the same solution set, they are equal for every value of c and d so x1 + kx2 = x1 + lx2 and I can reduce this down to k = l, what about c and d though? The problem statement tells me, I believe, c = d should I even show this?
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slider142
#2
Nov27-12, 04:57 PM
P: 898
Quote Quote by icesalmon View Post

3. The attempt at a solution
since the two equations have the same solution set, they are equal for every value of c and d so x1 + kx2 = x1 + lx2
That's not necessarily true. The two equations sharing the same solution set only means that if a and b satisfy a + kb = c, then a + lb = d. k, l, c and d are fixed numbers, so it is not clear what you mean by them being equal for every value of c and d. It is also not clear how this implies that a + kb = a + lb. The problem statement does not tell you that c = d; that is one of the two equalities you have to prove. Try a more direct approach. If a + kb = c, that gives you a fixed relationship between a and b.
icesalmon
#3
Nov27-12, 05:28 PM
P: 173
does a become dependent on b?

Mark44
#4
Nov27-12, 05:31 PM
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Question about linear equation

Quote Quote by icesalmon View Post
1. The problem statement, all variables and given/known data
Show that if the linear equations x1 + kx2 = c and x1 + lx2 = d have the same solution set, then the equations are identical ( k = l, c = d)

3. The attempt at a solution
since the two equations have the same solution set, they are equal for every value of c and d so x1 + kx2 = x1 + lx2 and I can reduce this down to k = l, what about c and d though? The problem statement tells me, I believe, c = d should I even show this?
What do you mean by "they are equal"? An equation is not equal to another equation, but two equations can be equivalent if they both have the same solution set.
slider142
#5
Nov27-12, 05:43 PM
P: 898
Quote Quote by icesalmon View Post
does a become dependent on b?
You can phrase it that way, or you can make b dependent on the value of a. Either way, the only thing we know from what they gave us is that a + lb = d for all values of a and b that satisfy a + kb = c.
You can take two approaches from here. You can either use substitution of equal values to formulate an argument about the relationship between l and k, and c and d. That one can be a bit subtle.
Or you can use specific values of a and b to argue for the same relationships, since a and b are variables. For example, is it always possible for b = 0 to be part of the solution set? If so, what conclusion can we draw from the particular solution that includes b = 0?
icesalmon
#6
Nov27-12, 05:54 PM
P: 173
I thought the problem statement told me that c was equal, and I do mean equal, to d somehow. And I thought that would make it possible to set them equal to one another, so moving forward with the false idea that c = d I let the two equations be equal similar to how functions can be set equal if you are trying to find intersection points, values where they would be the same.
haruspex
#7
Nov27-12, 06:07 PM
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Quote Quote by icesalmon View Post
I thought the problem statement told me that c was equal, and I do mean equal, to d somehow. And I thought that would make it possible to set them equal to one another
No, it's asking you to prove they're equal. You need to show that both c=d and k=l, without assuming either to start with.
icesalmon
#8
Nov27-12, 06:32 PM
P: 173
Quote Quote by slider142 View Post
You can phrase it that way, or you can make b dependent on the value of a. Either way, the only thing we know from what they gave us is that a + lb = d for all values of a and b that satisfy a + kb = c.
You can take two approaches from here. You can either use substitution of equal values to formulate an argument about the relationship between l and k, and c and d. That one can be a bit subtle.
Or you can use specific values of a and b to argue for the same relationships, since a and b are variables. For example, is it always possible for b = 0 to be part of the solution set? If so, what conclusion can we draw from the particular solution that includes b = 0?
that a = d, and a = c. With regards to it always being possible for b = 0 to be an element of the solution set, I would say yes although I don't have any mathematical proof and that's not good enough. But i'm confused about the question, are you saying that in some cases it can be a solution and in others it can't? How can that be possible?
haruspex
#9
Nov27-12, 07:39 PM
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Quote Quote by icesalmon View Post
With regards to it always being possible for b = 0 to be an element of the solution set, I would say yes although I don't have any mathematical proof and that's not good enough.
It comes to this: given that x1 + kx2 = c, is there a solution for x1 when x2=0? What does the equation say when x2 = 0?
icesalmon
#10
Nov27-12, 08:52 PM
P: 173
x1 = c
x1 = d
haruspex
#11
Nov27-12, 08:55 PM
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Quote Quote by icesalmon View Post
x1 = c
Right. So there is guaranteed a solution when x2 = 0, and the same (x1, x2) must be a solution of the other equation.
icesalmon
#12
Nov27-12, 11:12 PM
P: 173
and I can do the same thing with x1; letting x1 = 0 I get kx2 = c and x2= (c/k) so I have the point (0, c/k) that I can plug into the other equation because the solution sets are the same which I can use because the problem says that, I can just go ahead and solve to get k = l, is this sound? Maybe my wording isn't correct but I think it's the same method.
haruspex
#13
Nov28-12, 12:44 AM
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Quote Quote by icesalmon View Post
and I can do the same thing with x1; letting x1 = 0 I get kx2 = c and x2= (c/k) so I have the point (0, c/k) that I can plug into the other equation because the solution sets are the same which I can use because the problem says that, I can just go ahead and solve to get k = l, is this sound? Maybe my wording isn't correct but I think it's the same method.
Sounds good to me.


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