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mayankdevnani

  • 2 years ago

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

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

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

  3. ash2326
    • 2 years ago
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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 \[c=-\frac{4}{3}\]

  4. hartnn
    • 2 years ago
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    ^^ exactly

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

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

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

  8. ParthKohli
    • 2 years ago
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    Yes.

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

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

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

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

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

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

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

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

  17. Zarkon
    • 2 years ago
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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 \[\frac{7}{3}=13\]

  18. Zarkon
    • 2 years ago
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    thus no solution

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

  20. mayankdevnani
    • 2 years ago
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    and then solve it

  21. mathslover
    • 2 years ago
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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 .

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

  23. mathslover
    • 2 years ago
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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.

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

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

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

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

  28. mayankdevnani
    • 2 years ago
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    step by step

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

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

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

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

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

  34. ghazi
    • 2 years ago
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    -4/3

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

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

  37. mathslover
    • 2 years ago
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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

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

  39. ghazi
    • 2 years ago
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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

  40. ghazi
    • 2 years ago
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    @mayankdevnani clear?

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

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

  43. ghazi
    • 2 years ago
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    neither do they overlap

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

  45. ghazi
    • 2 years ago
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    :)

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

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

  48. ghazi
    • 2 years ago
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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

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

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