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Callisto
 3 years ago
Consider the system:
x + 2z =1
x+2y+4z =1
2xy+3z=2
For all solutions {x,y,z}, find the range of k such that \(k(x^23) > yz \)
Callisto
 3 years ago
Consider the system: x + 2z =1 x+2y+4z =1 2xy+3z=2 For all solutions {x,y,z}, find the range of k such that \(k(x^23) > yz \)

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Nubeer
 3 years ago
Best ResponseYou've already chosen the best response.0hmm i can find values of x , y and z but don;t know what to do after wards with them

Nubeer
 3 years ago
Best ResponseYou've already chosen the best response.0lol sorry guess not much help i am

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5Wait, can you find the numerical values for x, y, z?

Nubeer
 3 years ago
Best ResponseYou've already chosen the best response.0sure , give me a moment.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what about first finding the solution set of the system using algebra matrix

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0Gaussian elimination method.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5That's what I can do and that's why I ask if you can find the numerical value for {x,y,z}

Nubeer
 3 years ago
Best ResponseYou've already chosen the best response.0k 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.

Nubeer
 3 years ago
Best ResponseYou've already chosen the best response.0if i solve simultaneously, i got z = y and x = 1 2z

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5Yes, I got the same too, but how can we utilize it to find range of k?

Nubeer
 3 years ago
Best ResponseYou've already chosen the best response.0so 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.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5\[k[(12z)^23]>z^2\]\[k>\frac{z^2}{(12z)^23}\]Hmm..

Nubeer
 3 years ago
Best ResponseYou've already chosen the best response.0hmm got same thing.. after this i did the long division.. but lol not possible way out

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5I'm thinking of differetiating the fraction to find the min/max/range..

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5Suppose wolf. is right, then we get \(\frac{z(z+1)}{(2z^2+2z+1)^2}\) Put that derivative = 0, z=0 or z=1

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5Oh wait.. Why would I differentiate that!!!

Nubeer
 3 years ago
Best ResponseYou've already chosen the best response.0hmm thought u were trying to get max and min.

Nubeer
 3 years ago
Best ResponseYou've already chosen the best response.0yeah.. i still feel we using wrong approach.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x + 2z =1[1]\] \[x+2y+4z =1[2]\] \[2xy+3z=2[3]\] In Equation [1] \[x=12z\] In Equation [2] \[12z+2y+4z=1\] \[2y+2z=0[4]\] In Equation [3] \[2(12z)y+3z=2\] \[24zy+3z=2\] \[yz=0\] \[y+z=0[5]\] Therefore: \[x=12z\] or \[x=1+2y\] \[2z=1x\] \[z=\frac{1x}{2}\] \[2y=x1\] \[y=\frac{x1}{2}\] Therefore: \[yz=\frac{(1x)(x1)}{4}\] In: \[k(x^2−3)>yz\] \[k(x^23)>\frac{(1x)^2}{4}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[4k(x^23)>(1x)^2\] \[4kx^212k+(12x+x^2)>0\] \[4kx^212k+12x+x^2>0\] \[(4k+1)x^22x(12k1)>0\] Using the discriminant: \[b^24ac>0\] \[4+4(12k1)(4k+1)>0\] Solve as quadratic and find the range of values for k.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5Almost there, but still not correct.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you just solve for x,y,and z (using a system of equations) then plug the variables back in?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5I've tried solving the system above...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I ended up with x=1, y=0, z=0. So, \[k(1 ^{2}3)>(0)(0)\] =>\[k(2)>0\] =>k<0

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5It shouldn't have a unique solution.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh. What level of math is this?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0k belongs to (  infinity , 1/4 ] union (1/6 , 0 ) ?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5No, almost there. Can you tell us how you get there?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait, how is k<0 unique? (inf, 0)

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0k belongs to (  infinity , 1/4 ] union (1/6 , 0 ] ?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0can I post here after 30 minutes please? I have to go for dinner ?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5Sure, take your time :)

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Well first of all I solved for x and then put it into the eqn and then took diff. conditions .

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0k belongs to (  infinity , 1/4 ] union [1/6 , 0 ] ?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5No.. diff = different/differentiate?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0I am getting diff. answers. Wait!

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5I have some patience, so I keep waiting, and trying :)

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5Very very close!!! How do you get it??

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1(x1)^2 /(4x^2 12) tried to work with this :

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0k belongs to [1/6,0] union (infinity ,1/4] .. i am still getting it

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1Did you get what I did ?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1Coz I am not very confident myself! :P

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5I didn't get what you did :(

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5Did you start from this:\[k(x^23)>\frac{(1x)^2}{4}\]?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1I 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) ?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5The answer is correct, but I don't understand your explanation.

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0But aren't 0 and 1/6 satisfying the equation ?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5The inequality itself is > , not >=

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0k(x^2  3)>yz Put, x = 12z 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 ....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol x=1 btw I was having dinner z=0 and y =1/2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do 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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0And 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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0So k =1/6 also satisifies your inequality

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Well if z = 0 then go for it: k < 1/2

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5@aajugdar I don't mind if you check my matrix work above. I can't find a unique solution to the system.

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Well the solutions given by you and shubhamsrg have many cases left out .....

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5But when z=0, the inequality fails.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0put them in equation dude,the values are correct ok hold on,i wil tell ya

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0perform 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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Well ok in that case we get (by putting in my quadratic inequality in terms of z and k), k < 1/2 What about that ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0solve them,u will get values of x and z,from that values,find value of y by putting it in equation 2

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5@mathslover That is not even the answer @aajugdar Check this dw:1363447547599:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what operation did you perform on 3rd row????

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5@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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nope 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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0@Callisto very correct

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0remember 3 2 unknowns,2 equations,3 unknowns 3 eqautions.

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0(4k+1)z^2  4kz  2k>0 I used that (please have a look at my solution)

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5I am not a man This system has no unique solution, that's what the determinant of coefficient matrix tell us.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well,solve it correctly,it has a solution I found it

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5@aajugdar You can try using WolframAlpha, you can't get a unique solution. @mathslover Can you guarantee that (4k+1)>0?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1k > (x1)^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

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1only in that range x>f(x)

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Case  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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Case 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)

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Case  3: 4k+1 = 0 => k = 1/4 So, inequality reduces to z>1/2

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Which obviously consitutes part of real values of z.

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0So, 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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5I'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.

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0I hope I didn't annoy you?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5Nope, it just the exchange of ideas.

stamp
 3 years ago
Best ResponseYou've already chosen the best response.0it seems the solutions are not numbers, but rather variables (see attachment) also, http://tutorial.math.lamar.edu/Classes/Alg/AugmentedMatrixII.aspx

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5The solutions to the system are not numbers, that is it has no unique solution. The link doesn't wok.

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0@stamp read the question again ;)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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 GaussianElimination 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

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.51. 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So does the system have infinite solutions or a finite number of solutions?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5Infinitely many solution.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If it has infinitely many solutions then which solution gives the smallest value for yz / x^23?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I would say range is all real numbers?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no? do you know the answer then?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5I don't know how to get the answer though.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do 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.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5@genius12 Would you mind checking me steps of solving the system?dw:1363507811134:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did you do this through gaussian elimination?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Have 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5x + 2z =1 (1) x+2y+4z =1  (2) 2xy+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?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5If it is not infinitely many solutions, then it will be no solution.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5You'd better sleep now. Good night.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then there is infinitely many solutions.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wouldn't the system of equations be inconsistent?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5The system should be consistent.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, for you to solve the inequality and get the range for k, it should be consistent which it isn't.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@genius12 yes i agree it is inconsistent.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5Please point out my mistakes then.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5FYI: http://www.wolframalpha.com/input/?i=x+%2B+2z+%3D1+and+x%2B2y%2B4z+%3D1+and+2xy%2B3z%3D2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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. =.=

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5That's why we need to think how to get a way to solve it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Something wrong with the question? lol

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5No way! It is a past exam question.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What unit does this question cover?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5Matrix and determinant.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If it's a past exam, isn't there a given solution?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5The final answer  Yes.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5I know what the final answer is. I just don't know how to get the final answer.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.5When z=1, (max) => k=min (x, y, z) = (3, 1, 1) \[k>\frac{yz}{x^23}\]\[k>\frac{(1)(1)}{3^23}\]\[k>\frac{1}{6}\] When z = 0, (min) => k = max (x, y, z) = (1, 0, 0) \[k<\frac{yz}{x^23}\]\[k<\frac{(0)(0)}{1^23}\]\[k<0\] @shubhamsrg Your method works!! Too intuitive for me though :(

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1I 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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