anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Still trying to get through these review questions from what I missed while ill.
anonymous
  • anonymous
\[x_1-3x_2=1\\ 2x_1+hx_2=k\]?
anonymous
  • anonymous
Yep, 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.

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More answers

anonymous
  • anonymous
i got it
anonymous
  • anonymous
if you pick some random \(h\) and \(k\) there will probably be one unique solution
anonymous
  • anonymous
if you multiply the first equation by 2 you get \[2x_1-6x_2=2\\ 2x_2+hx_2=k\]
anonymous
  • anonymous
if you make \(h=-6\) and \(k=2\) they will be the same line exactly so there will be infinitely many solutions
anonymous
  • anonymous
if you make \(h=-6\) and \(k=\) anything but 2, they will be parallel lines with no intersection
anonymous
  • anonymous
Oh wow! How did you figure that out?
anonymous
  • anonymous
i am the satellite
anonymous
  • anonymous
ok 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
  • anonymous
Hehehe, that makes sense.
anonymous
  • anonymous
Thank you!

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