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Callisto

Consider the system: x + 2z =1 x+2y+4z =1 2x-y+3z=2 For all solutions {x,y,z}, find the range of k such that \(k(x^2-3) > yz \)

  • one year ago
  • one year ago

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  1. abb0t
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    I don't know!! :'(

    • one year ago
  2. nubeer
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    hmm i can find values of x , y and z but don;t know what to do after wards with them

    • one year ago
  3. Callisto
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    Same here...

    • one year ago
  4. nubeer
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    lol sorry guess not much help i am

    • one year ago
  5. Callisto
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    Wait, can you find the numerical values for x, y, z?

    • one year ago
  6. nubeer
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    sure , give me a moment.

    • one year ago
  7. Jonask
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    what about first finding the solution set of the system using algebra matrix

    • one year ago
  8. abb0t
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    Gaussian elimination method.

    • one year ago
  9. Callisto
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    |dw:1363361660822:dw|

    • one year ago
  10. Callisto
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    |dw:1363361796138:dw|

    • one year ago
  11. Callisto
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    That's what I can do and that's why I ask if you can find the numerical value for {x,y,z}

    • one year ago
  12. nubeer
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    k honestly i have done sole me this kind before where we get a who row zero but hard to explain the logic behind it..that whole zero row just made hard to find definite answer.. i normally would do this with the simultaneous method now. so we can get x, y , and z in terms of each other.

    • one year ago
  13. nubeer
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    if i solve simultaneously, i got z = -y and x = 1 -2z

    • one year ago
  14. Callisto
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    Yes, I got the same too, but how can we utilize it to find range of k?

    • one year ago
  15. nubeer
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    |dw:1363362152973:dw|

    • one year ago
  16. nubeer
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    so i think we can get value of x , y and z if we suppose any value of z .. according to it , we will get other values.

    • one year ago
  17. Callisto
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    \[k[(1-2z)^2-3]>-z^2\]\[k>\frac{-z^2}{(1-2z)^2-3}\]Hmm..

    • one year ago
  18. nubeer
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    hmm got same thing.. after this i did the long division.. but lol not possible way out

    • one year ago
  19. Callisto
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    I'm thinking of differetiating the fraction to find the min/max/range..

    • one year ago
  20. nubeer
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    can try tht

    • one year ago
  21. Callisto
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    Suppose wolf. is right, then we get \(\frac{z(z+1)}{(-2z^2+2z+1)^2}\) Put that derivative = 0, z=0 or z=-1

    • one year ago
  22. Callisto
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    Oh wait.. Why would I differentiate that!!!

    • one year ago
  23. nubeer
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    hmm thought u were trying to get max and min.

    • one year ago
  24. Callisto
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    But it won't work?

    • one year ago
  25. nubeer
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    yeah.. i still feel we using wrong approach.

    • one year ago
  26. Azteck
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    \[x + 2z =1[1]\] \[x+2y+4z =1[2]\] \[2x-y+3z=2[3]\] In Equation [1] \[x=1-2z\] In Equation [2] \[1-2z+2y+4z=1\] \[2y+2z=0[4]\] In Equation [3] \[2(1-2z)-y+3z=2\] \[2-4z-y+3z=2\] \[-y-z=0\] \[y+z=0[5]\] Therefore: \[x=1-2z\] or \[x=1+2y\] \[2z=1-x\] \[z=\frac{1-x}{2}\] \[2y=x-1\] \[y=\frac{x-1}{2}\] Therefore: \[yz=\frac{(1-x)(x-1)}{4}\] In: \[k(x^2−3)>yz\] \[k(x^2-3)>-\frac{(1-x)^2}{4}\]

    • one year ago
  27. Azteck
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    \[4k(x^2-3)>-(1-x)^2\] \[4kx^2-12k+(1-2x+x^2)>0\] \[4kx^2-12k+1-2x+x^2>0\] \[(4k+1)x^2-2x-(12k-1)>0\] Using the discriminant: \[b^2-4ac>0\] \[4+4(12k-1)(4k+1)>0\] Solve as quadratic and find the range of values for k.

    • one year ago
  28. Callisto
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    Almost there, but still not correct.

    • one year ago
  29. mathslover
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    Man at work.

    • one year ago
  30. AravindG
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    am I late? :)

    • one year ago
  31. mathslover
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    I dont' think so.

    • one year ago
  32. linshan789
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    Can you just solve for x,y,and z (using a system of equations) then plug the variables back in?

    • one year ago
  33. Callisto
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    I've tried solving the system above...

    • one year ago
  34. linshan789
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    I ended up with x=1, y=0, z=0. So, \[k(1 ^{2}-3)>(0)(0)\] =>\[k(-2)>0\] =>k<0

    • one year ago
  35. Callisto
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    It shouldn't have a unique solution.

    • one year ago
  36. linshan789
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    Oh. What level of math is this?

    • one year ago
  37. Callisto
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    High school

    • one year ago
  38. mathslover
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    k belongs to ( - infinity , -1/4 ] union (-1/6 , 0 ) ?

    • one year ago
  39. Callisto
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    No, almost there. Can you tell us how you get there?

    • one year ago
  40. linshan789
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    wait, how is k<0 unique? (-inf, 0)

    • one year ago
  41. mathslover
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    k belongs to ( - infinity , -1/4 ] union (-1/6 , 0 ] ?

    • one year ago
  42. Callisto
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    No....

    • one year ago
  43. mathslover
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    can I post here after 30 minutes please? I have to go for dinner ?

    • one year ago
  44. Callisto
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    Sure, take your time :)

    • one year ago
  45. mathslover
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    I am back to work.

    • one year ago
  46. Callisto
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    Any guideline?

    • one year ago
  47. mathslover
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    Well first of all I solved for x and then put it into the eqn and then took diff. conditions .

    • one year ago
  48. mathslover
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    k belongs to ( - infinity , -1/4 ] union [-1/6 , 0 ] ?

    • one year ago
  49. Callisto
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    No.. diff = different/differentiate?

    • one year ago
  50. mathslover
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    different.

    • one year ago
  51. aajugdar
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    looks easy wait

    • one year ago
  52. mathslover
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    I am getting diff. answers. Wait!

    • one year ago
  53. Callisto
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    I have some patience, so I keep waiting, and trying :)

    • one year ago
  54. mathslover
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    That is good.

    • one year ago
  55. shubhamsrg
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    is it [0, -1/6] ?

    • one year ago
  56. Callisto
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    Very very close!!! How do you get it??

    • one year ago
  57. shubhamsrg
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    -(x-1)^2 /(4x^2 -12) tried to work with this :|

    • one year ago
  58. mathslover
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    k belongs to [-1/6,0] union (-infinity ,1/4] .. i am still getting it

    • one year ago
  59. mathslover
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    is it wrong^ ?

    • one year ago
  60. mathslover
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    @Callisto ?

    • one year ago
  61. Callisto
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    It is wrong.

    • one year ago
  62. shubhamsrg
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    Did you get what I did ?

    • one year ago
  63. shubhamsrg
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    Coz I am not very confident myself! :P

    • one year ago
  64. Callisto
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    I didn't get what you did :(

    • one year ago
  65. Callisto
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    Did you start from this:\[k(x^2-3)>-\frac{(1-x)^2}{4}\]?

    • one year ago
  66. shubhamsrg
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    I am calling that f(x) k> f(x) whatever be the range of f(x), k's range will be opposite of that, i.e. maxima of f(x) is min of k and vice versa. so, is the ans (0,-1/6) ?

    • one year ago
  67. Callisto
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    The answer is correct, but I don't understand your explanation.

    • one year ago
  68. mathslover
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    But aren't 0 and -1/6 satisfying the equation ?

    • one year ago
  69. Callisto
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    The inequality itself is > , not >=

    • one year ago
  70. mathslover
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    k(x^2 - 3)>yz Put, x = 1-2z and y = -z Then k{1+4z^2 -4z - 3} > -z^2 => k(4z^2 - 4z - 2) > -z^2 => (4k+1)z^2 - 4kz - 2k>0 Now put, k = 0 you get z^2 > 0 Which is true ...... So k = 0 satisfies your inequality ....

    • one year ago
  71. mathslover
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    @Callisto

    • one year ago
  72. aajugdar
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    lol x=1 btw I was having dinner z=0 and y =-1/2

    • one year ago
  73. aajugdar
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    do this, eliminate 1 from lowest row in matrix you get two equations of x and z,solve them then u wil get answers of x,y,z

    • one year ago
  74. mathslover
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    And MY z = -1/6 have a look: (4k+1)z^2 - 4kz - 2k>0 Put k =-1/6 We get z^2 + 2z + 1 >0 OOPS that is true toooooo

    • one year ago
  75. mathslover
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    So k =-1/6 also satisifies your inequality

    • one year ago
  76. Callisto
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    What if z=0?

    • one year ago
  77. mathslover
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    Well if z = 0 then go for it: k < 1/2

    • one year ago
  78. aajugdar
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    yes z=0

    • one year ago
  79. Callisto
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    @aajugdar I don't mind if you check my matrix work above. I can't find a unique solution to the system.

    • one year ago
  80. mathslover
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    Well the solutions given by you and shubhamsrg have many cases left out .....

    • one year ago
  81. Callisto
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    But when z=0, the inequality fails.

    • one year ago
  82. aajugdar
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    put them in equation dude,the values are correct ok hold on,i wil tell ya

    • one year ago
  83. aajugdar
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    perform R3- R2/2 on R3 and then again arrange them in equation form,u will get x+2z =1 and then 3/2x+z =3/2

    • one year ago
  84. mathslover
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    Well ok in that case we get (by putting in my quadratic inequality in terms of z and k), k < 1/2 What about that ?

    • one year ago
  85. aajugdar
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    solve them,u will get values of x and z,from that values,find value of y by putting it in equation 2

    • one year ago
  86. Callisto
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    @mathslover That is not even the answer @aajugdar Check this |dw:1363447547599:dw|

    • one year ago
  87. aajugdar
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    what operation did you perform on 3rd row????

    • one year ago
  88. Callisto
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    -2 x R1 + R3 -> R3

    • one year ago
  89. Callisto
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    @mathlover When I learnt inequality, my teacher told me that if (1) ax^2+bx+c>0, where a>0, then the solution would be like (alpha)<x or x>(beta) (2) ax^2+bx+c<0, where a>0, then the solution would be like (alpha)<x<(beta)

    • one year ago
  90. aajugdar
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    nope man you didn't solve it as it should be solved,you see by makng complete 1 row 0 you wil never get solution you shouldn't keep a row NULL its a rule

    • one year ago
  91. mathslover
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    @Callisto very correct

    • one year ago
  92. aajugdar
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    remember 3 2 unknowns,2 equations,3 unknowns 3 eqautions.

    • one year ago
  93. mathslover
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    (4k+1)z^2 - 4kz - 2k>0 I used that (please have a look at my solution)

    • one year ago
  94. Callisto
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    I am not a man This system has no unique solution, that's what the determinant of coefficient matrix tell us.

    • one year ago
  95. aajugdar
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    2***

    • one year ago
  96. aajugdar
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    well,solve it correctly,it has a solution I found it

    • one year ago
  97. Callisto
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    @aajugdar You can try using WolframAlpha, you can't get a unique solution. @mathslover Can you guarantee that (4k+1)>0?

    • one year ago
  98. shubhamsrg
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    k > -(x-1)^2 /(4x^2 -12) the derivatives are 0 at x=3 and 1 respective values of y are -1/6 (local maxima) and 0(local minima) so k > -1/6 and k<0

    • one year ago
  99. aajugdar
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    -facewall-

    • one year ago
  100. shubhamsrg
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    only in that range x>f(x)

    • one year ago
  101. shubhamsrg
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    I mean k>f(x)

    • one year ago
  102. mathslover
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    Case - 1 : 4k + 1 > 0 => k>-1/4 Now, that means the inequality has a graph upward. Now for the inequality >0. b^2 - 4ac < 0 => k(6k+1) < 0 => If k< 0 then 6k+1>0 i.e. k>-1/6 And if k>0 then 6k+1 < 0 i.e. k<-1/6 (This option is not possible) So, k<0 ; k>-1/6 ; and k>-1/4 Take intersection: k belongs to (-1/6,0) @Callisto I am thus taking cases

    • one year ago
  103. mathslover
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    Case -2 : (4k+1)<0 => k <-1/4 i .e. the graph is bent downwards. Now for inequality> 0. b^2 - 4ac > 0 => k(6k+1) >0 => k> 0 and k>-1/6 or k<0 and k<-1/6 => k>0 or k<-1/6 But we took k<-1/4 So, k>0 case is eliminated. So, k belongs to (-infinity, -1/4)

    • one year ago
  104. mathslover
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    Case - 3: 4k+1 = 0 => k = -1/4 So, inequality reduces to z>-1/2

    • one year ago
  105. mathslover
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    Which obviously consitutes part of real values of z.

    • one year ago
  106. mathslover
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    So, I said answer should be: k belongs to (-infinity, -1/4] union (-1/6,0)........ Kindly please tell if I am correct or If I went wrong somewere

    • one year ago
  107. mathslover
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    Well I agree with you that k = -1/4 and -1/6 do not constitute the solution so modify it k belongs to (-infinity, -1/4) union (-1/6,0) But I still don't know that how will I then deal with 4k+1 = 0 case

    • one year ago
  108. Callisto
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    I'm thinking.. for the inequality, (4k+1)z^2 - 4kz - 2k>0, it doesn't imply that (4k+1) >0 or (4k+1)<0. It can be 4k+1 =0 Oh, you've pointed the problem out.

    • one year ago
  109. mathslover
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    Think about that

    • one year ago
  110. mathslover
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    I hope I didn't annoy you?

    • one year ago
  111. Callisto
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    Nope, it just the exchange of ideas.

    • one year ago
  112. Callisto
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    *it's

    • one year ago
  113. stamp
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    it seems the solutions are not numbers, but rather variables (see attachment) also, http://tutorial.math.lamar.edu/Classes/Alg/AugmentedMatrixII.aspx

    • one year ago
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  114. Callisto
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    The solutions to the system are not numbers, that is it has no unique solution. The link doesn't wok.

    • one year ago
  115. Callisto
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    *work

    • one year ago
  116. mathslover
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    @stamp read the question again ;)

    • one year ago
  117. genius12
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    The system is easy to solve for x, y, z. It's just a little tedious. I am in a rush as it's nearly 4 am so I'll give you a brief solution to this problem. First rearrange x + 2z = 1 to z = (1 - x) / 2 Now you have z in terms of x. Find the value of y in terms of x from the second equation and plug in (1 - x)/2 for z in it. Then plug these y and z values in terms of x in the third equation and solve for x. Then solve for z and then y. You can of course use Gaussian-Elimination Method but that can be annoying at times. Anyway, after you got the solution for x,y,z, just plug the values in to the inequality and rearrange it to k > yz/(x^2 - 3). And that will give you the range for k. G2g now. @Callisto

    • one year ago
  118. Callisto
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    1. There is not unique solution to the system. I have solved it using Gaussian elimination and have already put what I have got above. 2. I have put x, y, z into the inequality (in terms of z), but it just leaves me a range with unknown z. 3. I have tried differentiate z to get min/max, but the values I got (z=0/z=-1) does not match with the answer.

    • one year ago
  119. genius12
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    So does the system have infinite solutions or a finite number of solutions?

    • one year ago
  120. Callisto
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    Infinitely many solution.

    • one year ago
  121. genius12
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    If it has infinitely many solutions then which solution gives the smallest value for yz / x^2-3?

    • one year ago
  122. some_someone
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    I would say range is all real numbers?

    • one year ago
  123. Callisto
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    No.

    • one year ago
  124. some_someone
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    no? do you know the answer then?

    • one year ago
  125. Callisto
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    I know the answer.

    • one year ago
  126. Callisto
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    I don't know how to get the answer though.

    • one year ago
  127. some_someone
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    oh ok got you

    • one year ago
  128. genius12
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    Do you know what the smallest solution is for x,y,z such that yz / x^2 - 3 is the smallest possible value? Try to find such a solution.

    • one year ago
  129. Callisto
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    @genius12 Would you mind checking me steps of solving the system?|dw:1363507811134:dw|

    • one year ago
  130. Callisto
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    *my

    • one year ago
  131. genius12
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    Did you do this through gaussian elimination?

    • one year ago
  132. Callisto
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    Yes.

    • one year ago
  133. genius12
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    Have you tried solving this algebraically? It will be tedious, but could be promising. It's easy to mess up on matrices, but not as much when you solve algebraically.

    • one year ago
  134. genius12
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    I know for sure that you will be able to solve this algebraically, and I am pretty confident that you won't get "infinitely many solutions", hopefully. 4:16 AM. Dark room, people sleeping in here, I think I'll head to sleep as well. Good luck Callisto.

    • one year ago
  135. Callisto
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    x + 2z =1 ---(1) x+2y+4z =1 --- (2) 2x-y+3z=2 ---(3) -(1)+(2) 2y + 2z = 0 => y = -z --- (4) Sub. (4) into (3) 2x - (-z) + 3z =2 x+2z = 1 --- (5) , which is also (1) Pretty obvious, huh?

    • one year ago
  136. Callisto
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    If it is not infinitely many solutions, then it will be no solution.

    • one year ago
  137. Callisto
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    You'd better sleep now. Good night.

    • one year ago
  138. genius12
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    Then there is infinitely many solutions.

    • one year ago
  139. Callisto
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    What's next?

    • one year ago
  140. some_someone
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    Wouldn't the system of equations be inconsistent?

    • one year ago
  141. Callisto
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    The system should be consistent.

    • one year ago
  142. genius12
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    Yes, for you to solve the inequality and get the range for k, it should be consistent which it isn't.

    • one year ago
  143. some_someone
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    @genius12 yes i agree it is inconsistent.

    • one year ago
  144. Callisto
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    Please point out my mistakes then.

    • one year ago
  145. Callisto
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    FYI: http://www.wolframalpha.com/input/?i=x+%2B+2z+%3D1+and+x%2B2y%2B4z+%3D1+and+2x-y%2B3z%3D2

    • one year ago
  146. genius12
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    I could pick any value for z and get the same three equations. This gives us no way to come up with a range for k. =.=

    • one year ago
  147. Callisto
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    That's why we need to think how to get a way to solve it.

    • one year ago
  148. genius12
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    Something wrong with the question? lol

    • one year ago
  149. Callisto
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    No way! It is a past exam question.

    • one year ago
  150. genius12
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    What unit does this question cover?

    • one year ago
  151. Callisto
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    Matrix and determinant.

    • one year ago
  152. genius12
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    If it's a past exam, isn't there a given solution?

    • one year ago
  153. Callisto
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    The final answer - Yes.

    • one year ago
  154. genius12
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    Check it? lol

    • one year ago
  155. Callisto
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    I know what the final answer is. I just don't know how to get the final answer.

    • one year ago
  156. Callisto
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    When z=-1, (max) => k=min (x, y, z) = (3, 1, -1) \[k>\frac{yz}{x^2-3}\]\[k>\frac{(1)(-1)}{3^2-3}\]\[k>-\frac{1}{6}\] When z = 0, (min) => k = max (x, y, z) = (1, 0, 0) \[k<\frac{yz}{x^2-3}\]\[k<\frac{(0)(0)}{1^2-3}\]\[k<0\] @shubhamsrg Your method works!! Too intuitive for me though :(

    • one year ago
  157. shubhamsrg
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    I am not too sure how to put my words more properly. :| Hey, why don't you try to understand this with the graph, you may use wolfram for the graphing.

    • one year ago
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