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anonymous
 one year ago
Choose h and k such that the system has (a) no solution (b) a unique solution or (c) many solutions. Give separate answers for each part. x1  3x2 = 1 and 2x1 + hx2 = k
anonymous
 one year ago
Choose h and k such that the system has (a) no solution (b) a unique solution or (c) many solutions. Give separate answers for each part. x1  3x2 = 1 and 2x1 + hx2 = k

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Still trying to get through these review questions from what I missed while ill.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x_13x_2=1\\ 2x_1+hx_2=k\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, those are the correct systems. I should have made sure to mention that they were row labels and not exponents. I'm sorry if that was confusing.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you pick some random \(h\) and \(k\) there will probably be one unique solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you multiply the first equation by 2 you get \[2x_16x_2=2\\ 2x_2+hx_2=k\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you make \(h=6\) and \(k=2\) they will be the same line exactly so there will be infinitely many solutions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you make \(h=6\) and \(k=\) anything but 2, they will be parallel lines with no intersection

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh wow! How did you figure that out?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok in truth if you wanted to solve it you would have to match up the coefficients to use "elimination" i just matched up the coefficients of the \(x_1\) term by doubling the first one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hehehe, that makes sense.
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