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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

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  1. anonymous
    • one year ago
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    Still trying to get through these review questions from what I missed while ill.

  2. anonymous
    • one year ago
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    \[x_1-3x_2=1\\ 2x_1+hx_2=k\]?

  3. anonymous
    • one year ago
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    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.

  4. anonymous
    • one year ago
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    i got it

  5. anonymous
    • one year ago
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    if you pick some random \(h\) and \(k\) there will probably be one unique solution

  6. anonymous
    • one year ago
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    if you multiply the first equation by 2 you get \[2x_1-6x_2=2\\ 2x_2+hx_2=k\]

  7. anonymous
    • one year ago
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    if you make \(h=-6\) and \(k=2\) they will be the same line exactly so there will be infinitely many solutions

  8. anonymous
    • one year ago
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    if you make \(h=-6\) and \(k=\) anything but 2, they will be parallel lines with no intersection

  9. anonymous
    • one year ago
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    Oh wow! How did you figure that out?

  10. anonymous
    • one year ago
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    i am the satellite

  11. anonymous
    • one year ago
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    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

  12. anonymous
    • one year ago
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    Hehehe, that makes sense.

  13. anonymous
    • one year ago
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    Thank you!

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