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mariomintchev

  • one year ago

5x+5y=1 25x+ky=4 solve for k

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  1. Kuoministers
    • one year ago
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    there is few ways of working this out. Elimination, Substitution and maybe multiplication

  2. Kuoministers
    • one year ago
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    which way do you prefer?

  3. mariomintchev
    • one year ago
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    i need my mind refreshed. whichever method works best is the method i choose.

  4. Kuoministers
    • one year ago
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    ill use substitution then :P x = \[x = y - \frac{ 1 }{ 5 }\] 25x + ky = 4

  5. Kuoministers
    • one year ago
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    25(y - \[25(y - \frac{ 1 }{ 5 }) + ky = 4\]

  6. Kuoministers
    • one year ago
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    \[25y - 5 + ky = 4\]

  7. Kuoministers
    • one year ago
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    do you get it now?

  8. mariomintchev
    • one year ago
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    howd u get y-(1/5)?

  9. Kuoministers
    • one year ago
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    divide everything in the first equation by 5

  10. Kuoministers
    • one year ago
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    there is no solution :)

  11. Kuoministers
    • one year ago
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    ill use elimination now

  12. Kuoministers
    • one year ago
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    times the first equation by -5 -25x - 25y = -5 25x + ky = 4

  13. Kuoministers
    • one year ago
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    now add this equation vertically together. it would give you k=25

  14. mariomintchev
    • one year ago
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    i probably shoulda phrased the question the way they have it on my hw. it says: Determine the value of k for which the system of linear equations has no solutions.

  15. Kuoministers
    • one year ago
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    haha great :/ u say that now

  16. Kuoministers
    • one year ago
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    you know why there is no solution? if you sub K = 25 it wont work

  17. mariomintchev
    • one year ago
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    :(

  18. mariomintchev
    • one year ago
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    yes?

  19. mariomintchev
    • one year ago
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    because they equal different things

  20. Kuoministers
    • one year ago
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    \[\Huge{\color{Purple}{\\{\ \boxed{\mathbb{\text{}{}}{\color{green}{\color{blue}T{\color{purple}h{\color{pink}a{\color{red}nk}}}}}}}}} \] \[\Huge{\color{Purple}{\\{\ \boxed{\mathbb{\text{}{}}{\color{blue}Y{\color{green}o{\color{red}u}}}}}}} \]

  21. mariomintchev
    • one year ago
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    haha no, THANK YOU ! :)

  22. mariomintchev
    • one year ago
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    ok i have another one lol this one wants the k to have an infinite amount of solutions

  23. mariomintchev
    • one year ago
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    49x-49y=49 7x+ky=7

  24. mariomintchev
    • one year ago
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    the answer is not 1,7,49,1/7, or 1/49.

  25. mariomintchev
    • one year ago
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    @Kuoministers

  26. Kuoministers
    • one year ago
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    sorry been bit busy

  27. Kuoministers
    • one year ago
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    so do you have an idea where to start?

  28. Kuoministers
    • one year ago
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    i would times the 2nd equation by 7

  29. mariomintchev
    • one year ago
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    ok...

  30. Kuoministers
    • one year ago
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    do you undertand this yet lol?

  31. Kuoministers
    • one year ago
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    49x-49y=49 7x+ky=7 49x-49y=49 49x + 7ky = 49

  32. mariomintchev
    • one year ago
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    dont know where to go from there

  33. Kuoministers
    • one year ago
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    -49y - 7ky = 0 -7y(7 + k) = 0

  34. mariomintchev
    • one year ago
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    how do we get the k from that

  35. Kuoministers
    • one year ago
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    im bit confused :P you sure their is more than 1 answer??

  36. mariomintchev
    • one year ago
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    its 1 answer

  37. Kuoministers
    • one year ago
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    oh u said this one wants the k to have an infinite amount of solutions

  38. mariomintchev
    • one year ago
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    1 K but infinite solutions we need the K

  39. Kuoministers
    • one year ago
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    y(7 + k ) = 0 from this can you find K? im pretty sure you can :P

  40. Kuoministers
    • one year ago
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    hint: you only need to look at this question to get K

  41. mariomintchev
    • one year ago
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    ok its -7 lol

  42. mariomintchev
    • one year ago
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    im feeling slooow ha

  43. Kuoministers
    • one year ago
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    ok i g2g now :)

  44. Kuoministers
    • one year ago
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    have fun with your maths

  45. mariomintchev
    • one year ago
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    haha see ya.

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