
#1
Nov2712, 04:43 PM

P: 173

1. The problem statement, all variables and given/known data
Show that if the linear equations x_{1} + kx_{2} = c and x_{1} + lx_{2} = 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 x_{1} + kx_{2} = x_{1} + lx_{2} 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? 



#2
Nov2712, 04:57 PM

P: 876





#3
Nov2712, 05:28 PM

P: 173

does a become dependent on b?




#4
Nov2712, 05:31 PM

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P: 20,933

question about linear equation 



#5
Nov2712, 05:43 PM

P: 876

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? 



#6
Nov2712, 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.




#7
Nov2712, 06:07 PM

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#8
Nov2712, 06:32 PM

P: 173





#9
Nov2712, 07:39 PM

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#10
Nov2712, 08:52 PM

P: 173

x_{1} = c
x_{1} = d 



#11
Nov2712, 08:55 PM

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#12
Nov2712, 11:12 PM

P: 173

and I can do the same thing with x_{1}; letting x_{1} = 0 I get kx_{2} = c and x_{2}= (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.




#13
Nov2812, 12:44 AM

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