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mayankdevnani Group Title

Given: 3x-4y=7 and x+cy=13 , for what value of "c" will the two equations not have a solution?

  • 2 years ago
  • 2 years ago

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  1. mayankdevnani Group Title
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    options are : a) \[\frac{3}{4}\] b)\[\frac{4}{3}\] c) \[-4\] d) \[\frac{-4}{3}\]

    • 2 years ago
  2. hartnn Group Title
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    i think there is easier way out....

    • 2 years ago
  3. ash2326 Group Title
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    Rewrite first equation \[x-\frac{4}{3}y=7\] for no solution we need to have parallel lines second equation \[x+cy=13\] compare this with the first \[c=-\frac{4}{3}\]

    • 2 years ago
  4. hartnn Group Title
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    ^^ exactly

    • 2 years ago
  5. ParthKohli Group Title
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    @hartnn: Oh yeah

    • 2 years ago
  6. sauravshakya Group Title
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    WELL FOLLOW @ash2326 THAT ONE MUCH EASIER..... agreed with @hartnn

    • 2 years ago
  7. hartnn Group Title
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    no solution is of the form: ax+b=c ax+b=d

    • 2 years ago
  8. ParthKohli Group Title
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    Yes.

    • 2 years ago
  9. mayankdevnani Group Title
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    @ash2326 gd and then

    • 2 years ago
  10. Zarkon Group Title
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    and why would that have an infinite # of solutions?

    • 2 years ago
  11. hartnn Group Title
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    because ax+b=c and ax+b=c are satisfied by infinite number of points (x,y) simultaneously.

    • 2 years ago
  12. Zarkon Group Title
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    ash2326's solution is mostly correct

    • 2 years ago
  13. Zarkon Group Title
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    this one doesn't satisfy that

    • 2 years ago
  14. mayankdevnani Group Title
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    ok! then @Zarkon

    • 2 years ago
  15. Zarkon Group Title
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    3x-4y=7 and x+cy=13 \[x-\frac{4}{3}y=\frac{7}{3}\] \[x+cy=13\]

    • 2 years ago
  16. mayankdevnani Group Title
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    yaa ... i also agree with @Zarkon don't you @mathslover

    • 2 years ago
  17. Zarkon Group Title
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    if c=-4/3 then you get \[x-\frac{4}{3}y=\frac{7}{3}\] and \[x-\frac{4}{3}y=13\] ie \[\frac{7}{3}=13\]

    • 2 years ago
  18. Zarkon Group Title
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    thus no solution

    • 2 years ago
  19. mayankdevnani Group Title
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    isn't like this \[3x-4y=7\] \[3x=7-4y\] \[x-y=\frac{7}{12}\] \[x=\frac{7}{12}+y\]

    • 2 years ago
  20. mayankdevnani Group Title
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    and then solve it

    • 2 years ago
  21. mathslover Group Title
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    What I did is this : \[\large{7+4y = 3(13-cy)}\] \[\large{7+4y=39-3cy}\] \[\large{32= 4y+3cy}\] We have to find the value of c when the RHS becomes zero which will give no solution. \[\large{4y+3(c)(y)=0}\] \[\large{c = \frac{-4y}{3y}}\] \[\large{c=\frac{-4}{3}}\] hence at c = -4/3 we get no solutions for the variables in the given two equations .

    • 2 years ago
  22. sauravshakya Group Title
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    @mayankdevnani what r u doing?

    • 2 years ago
  23. mathslover Group Title
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    @mayankdevnani you need to concentrate hardly on understanding questions... You are solving for x or y but in the question we have to find the value for c which makes the solutions for the equation as null set.

    • 2 years ago
  24. mayankdevnani Group Title
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    how did you get it......... @mathslover

    • 2 years ago
  25. mathslover Group Title
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    btw , @Zarkon sir , was my method acceptable? I know there are easy methods too.. but that is what I did..

    • 2 years ago
  26. mayankdevnani Group Title
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    ANSWER is correct... but i did'nt understand..... plz tell me once again.....

    • 2 years ago
  27. mathslover Group Title
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    @mayankdevnani where are you not getting ?

    • 2 years ago
  28. mayankdevnani Group Title
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    step by step

    • 2 years ago
  29. sauravshakya Group Title
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    THAT WAS exactly what I was thinking @mathslover THAT IS ABSOULETLY CORRECT

    • 2 years ago
  30. Zarkon Group Title
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    yes...your way is fine

    • 2 years ago
  31. mathslover Group Title
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    OK thanks @sauravshakya Let us wait for honorable zarkon sir ..

    • 2 years ago
  32. mathslover Group Title
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    OK thanks a lot @Zarkon .. that fine made a lot sense sir.. :)

    • 2 years ago
  33. mayankdevnani Group Title
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    @mathslover how did you get it \[7+4y=1(13-cy)\]

    • 2 years ago
  34. ghazi Group Title
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    -4/3

    • 2 years ago
  35. mayankdevnani Group Title
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    ya.. but how??

    • 2 years ago
  36. ghazi Group Title
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    that will be a prallal line

    • 2 years ago
  37. mathslover Group Title
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    3x-4y = 7 x = (7+4y) / 3 and then : x + cy = 13 x = 13-cy 13-cy = (7+4y)/3 (since they are equal to x) 3(13-cy) = 7+4y

    • 2 years ago
  38. mathslover Group Title
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    got it @mayankdevnani ?

    • 2 years ago
  39. ghazi Group Title
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    just put -4/3 at C and you will have a parallel set of line and parallel set of lines have no solution

    • 2 years ago
  40. ghazi Group Title
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    @mayankdevnani clear?

    • 2 years ago
  41. mayankdevnani Group Title
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    ok... @mathslover

    • 2 years ago
  42. ghazi Group Title
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    3x-4y=7 and 3x-4y=39 are parallel lines how could they have a solution ..they intersect nowhere

    • 2 years ago
  43. ghazi Group Title
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    neither do they overlap

    • 2 years ago
  44. mayankdevnani Group Title
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    thnx.... @Zarkon @ghazi @sauravshakya @ash2326 but special thnx to @mathslover

    • 2 years ago
  45. ghazi Group Title
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    :)

    • 2 years ago
  46. ghazi Group Title
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    @mathslover is my approach correct?

    • 2 years ago
  47. sauravshakya Group Title
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    Isnt your method same as @ash2326 's method........ @ghazi

    • 2 years ago
  48. ghazi Group Title
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    i didn't see ...i just came here read the question and identified these are parallel lines...i guess yes both have same solution and we don't need to solve anything because already it is visible that it's case of parallel line when c= -4/3 hence no solution

    • 2 years ago
  49. ghazi Group Title
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    yes it is similar to that of @ash2326

    • 2 years ago
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