No problem, glad I could help! :smile:

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Homework Statement


In the following problem, find conditions on a and b such that the system has no solution, one solution, and infinitely many solutions.

x + by = -1
ax + 2y = 5


Homework Equations


None that I know of.


The Attempt at a Solution


Basically, first I put the entire equation into a matrix.
[ 1 b | -1
a 2 | 5 ]
I reduce the bottom by subtracting R2 - aR1
[ 1 b | -1
0 2-ab| 5 +a ]
I then reduce the bottom again by dividing R2/(2-ab)
[ 1 b | -1
0 1 | (5+a)/(2-ab)]
I remove the b from the top by subtraction: R1 - bR2
[1 0 | -1 - b((5+a)/(2-ab))
0 1 | (5+a)/(2-ab) ]
This leaves me with the values for x and y, and for the first question I am correct in saying if ab = 2, then there is no solution as it is undefined. However, my unique solution is somehow wrong and I would like some help in determining if I made an error or I somehow didn't reduce something.

The correct unique solution is: x = (-2 - 5b)/(2-ab) y = (a+5)/(2-ab)

Also, I have no idea what finding an infinite solution means, I would really like some help on clarifying that. Thank you.
 
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Hi Temp0! Welcome to PF! :smile:
Temp0 said:
[1 0 | -1 - b((5+a)/(2-ab))
0 1 | (5+a)/(2-ab) ]

The correct unique solution is: x = (-2 - 5b)/(2-ab) y = (a+5)/(2-ab)

That is the same as yours! :wink:
Also, I have no idea what finding an infinite solution means

Hint: if A and B are two solutions, what can you say about A - B ? :smile:
 
Also, I have no idea what finding an infinite solution means, I would really like some help on clarifying that. Thank you.
For example: if you are given graphs of f(x) and g(x) [neither contain discontinuities] and g(x) 's graph never "falls under" or crosses f(x) , no matter what the argument, g(x) has always higher values than f(x) and you are asked to provide solutions for g(x) > f(x) then you can say that there are infinitely many solutions. But you are dealing with an equal sign so that must mean the 2 graphs are...?
 
tiny-tim said:
Hi Temp0! Welcome to PF! :smile:That is the same as yours! :wink:Hint: if A and B are two solutions, what can you say about A - B ? :smile:

Hmm, thanks for the help on the infinite thing, I finally get that ^^. However, I've tried reducing and expanding my answer, but it never becomes the same as the answer in the book. Are there any hints you can give me? =D

edit: nvm, I just looked at it again and realized how to get to the answer, thanks for your help =p.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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