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Callisto

  • one year ago

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 \(k(x^2-3) > yz \)

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  1. abb0t
    • one year ago
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    I don't know!! :'(

  2. nubeer
    • one year ago
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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

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

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

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

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

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

  8. abb0t
    • one year ago
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    Gaussian elimination method.

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

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

  11. Callisto
    • one year ago
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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}

  12. nubeer
    • one year ago
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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.

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

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

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

  16. nubeer
    • one year ago
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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.

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

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

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

  20. nubeer
    • one year ago
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    can try tht

  21. Callisto
    • one year ago
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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

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

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

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

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

  26. Azteck
    • one year ago
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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: \[k(x^2−3)>yz\] \[k(x^2-3)>-\frac{(1-x)^2}{4}\]

  27. Azteck
    • one year ago
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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.

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

  29. mathslover
    • one year ago
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    Man at work.

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

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

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

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

  34. linshan789
    • one year ago
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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

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

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

  37. Callisto
    • one year ago
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    High school

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

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

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

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

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

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

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

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

  46. Callisto
    • one year ago
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    Any guideline?

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

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

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

  50. mathslover
    • one year ago
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    different.

  51. aajugdar
    • one year ago
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    looks easy wait

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

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

  54. mathslover
    • one year ago
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    That is good.

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

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

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

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

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

  60. mathslover
    • one year ago
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    @Callisto ?

  61. Callisto
    • one year ago
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    It is wrong.

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

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

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

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

  66. shubhamsrg
    • one year ago
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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) ?

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

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

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

  70. mathslover
    • one year ago
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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 ....

  71. mathslover
    • one year ago
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    @Callisto

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

  73. aajugdar
    • one year ago
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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

  74. mathslover
    • one year ago
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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

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

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

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

  78. aajugdar
    • one year ago
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    yes z=0

  79. Callisto
    • one year ago
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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.

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

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

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

  83. aajugdar
    • one year ago
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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

  84. mathslover
    • one year ago
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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 ?

  85. aajugdar
    • one year ago
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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

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

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

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

  89. Callisto
    • one year ago
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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)

  90. aajugdar
    • one year ago
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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

  91. mathslover
    • one year ago
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    @Callisto very correct

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

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

  94. Callisto
    • one year ago
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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.

  95. aajugdar
    • one year ago
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    2***

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

  97. Callisto
    • one year ago
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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?

  98. shubhamsrg
    • one year ago
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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

  99. aajugdar
    • one year ago
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    -facewall-

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

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

  102. mathslover
    • one year ago
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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

  103. mathslover
    • one year ago
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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)

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

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

  106. mathslover
    • one year ago
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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

  107. mathslover
    • one year ago
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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

  108. Callisto
    • one year ago
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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.

  109. mathslover
    • one year ago
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    Think about that

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

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

  112. Callisto
    • one year ago
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    *it's

  113. stamp
    • one year ago
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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

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

  115. Callisto
    • one year ago
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    *work

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

  117. genius12
    • one year ago
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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

  118. Callisto
    • one year ago
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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.

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

  120. Callisto
    • one year ago
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    Infinitely many solution.

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

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

  123. Callisto
    • one year ago
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    No.

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

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

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

  127. some_someone
    • one year ago
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    oh ok got you

  128. genius12
    • one year ago
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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.

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

  130. Callisto
    • one year ago
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    *my

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

  132. Callisto
    • one year ago
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    Yes.

  133. genius12
    • one year ago
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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.

  134. genius12
    • one year ago
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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.

  135. Callisto
    • one year ago
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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?

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

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

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

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

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

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

  142. genius12
    • one year ago
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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.

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

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

  145. Callisto
    • one year ago
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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

  146. genius12
    • one year ago
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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. =.=

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

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

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

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

  151. Callisto
    • one year ago
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    Matrix and determinant.

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

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

  154. genius12
    • one year ago
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    Check it? lol

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

  156. Callisto
    • one year ago
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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 :(

  157. shubhamsrg
    • one year ago
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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.

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