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mayankdevnani

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

  • one year ago
  • one year ago

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

    • one year ago
  2. hartnn
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    i think there is easier way out....

    • one year ago
  3. ash2326
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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 \[c=-\frac{4}{3}\]

    • one year ago
  4. hartnn
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    ^^ exactly

    • one year ago
  5. ParthKohli
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    @hartnn: Oh yeah

    • one year ago
  6. sauravshakya
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    WELL FOLLOW @ash2326 THAT ONE MUCH EASIER..... agreed with @hartnn

    • one year ago
  7. hartnn
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    no solution is of the form: ax+b=c ax+b=d

    • one year ago
  8. ParthKohli
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    Yes.

    • one year ago
  9. mayankdevnani
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    @ash2326 gd and then

    • one year ago
  10. Zarkon
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    and why would that have an infinite # of solutions?

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

    • one year ago
  12. Zarkon
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    ash2326's solution is mostly correct

    • one year ago
  13. Zarkon
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    this one doesn't satisfy that

    • one year ago
  14. mayankdevnani
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    ok! then @Zarkon

    • one year ago
  15. Zarkon
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    3x-4y=7 and x+cy=13 \[x-\frac{4}{3}y=\frac{7}{3}\] \[x+cy=13\]

    • one year ago
  16. mayankdevnani
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    yaa ... i also agree with @Zarkon don't you @mathslover

    • one year ago
  17. Zarkon
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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 \[\frac{7}{3}=13\]

    • one year ago
  18. Zarkon
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    thus no solution

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

    • one year ago
  20. mayankdevnani
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    and then solve it

    • one year ago
  21. mathslover
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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 .

    • one year ago
  22. sauravshakya
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    @mayankdevnani what r u doing?

    • one year ago
  23. mathslover
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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.

    • one year ago
  24. mayankdevnani
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    how did you get it......... @mathslover

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

    • one year ago
  26. mayankdevnani
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    ANSWER is correct... but i did'nt understand..... plz tell me once again.....

    • one year ago
  27. mathslover
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    @mayankdevnani where are you not getting ?

    • one year ago
  28. mayankdevnani
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    step by step

    • one year ago
  29. sauravshakya
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    THAT WAS exactly what I was thinking @mathslover THAT IS ABSOULETLY CORRECT

    • one year ago
  30. Zarkon
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    yes...your way is fine

    • one year ago
  31. mathslover
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    OK thanks @sauravshakya Let us wait for honorable zarkon sir ..

    • one year ago
  32. mathslover
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    OK thanks a lot @Zarkon .. that fine made a lot sense sir.. :)

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

    • one year ago
  34. ghazi
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    -4/3

    • one year ago
  35. mayankdevnani
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    ya.. but how??

    • one year ago
  36. ghazi
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    that will be a prallal line

    • one year ago
  37. mathslover
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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

    • one year ago
  38. mathslover
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    got it @mayankdevnani ?

    • one year ago
  39. ghazi
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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

    • one year ago
  40. ghazi
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    @mayankdevnani clear?

    • one year ago
  41. mayankdevnani
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    ok... @mathslover

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

    • one year ago
  43. ghazi
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    neither do they overlap

    • one year ago
  44. mayankdevnani
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    thnx.... @Zarkon @ghazi @sauravshakya @ash2326 but special thnx to @mathslover

    • one year ago
  45. ghazi
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    :)

    • one year ago
  46. ghazi
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    @mathslover is my approach correct?

    • one year ago
  47. sauravshakya
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    Isnt your method same as @ash2326 's method........ @ghazi

    • one year ago
  48. ghazi
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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

    • one year ago
  49. ghazi
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    yes it is similar to that of @ash2326

    • one year ago
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