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ganeshie8

  • 2 years ago

There are two straight lines y1 = (3/7)x y2 = (2/7)x + 343 Find "At what POSITIVE value of x, y1 reaches twice of y2 " i am struggling with this problem from an hour... pls help using geometry only. no calculus pls.

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  1. shubhamsrg
    • 2 years ago
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    using geometry? maybe you mean using purely algebra?

  2. ganeshie8
    • 2 years ago
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    ya solving equations... algebra is ok.. . :)

  3. shubhamsrg
    • 2 years ago
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    just set y1 = 2y2 make substitutions and solve for x

  4. lgbasallote
    • 2 years ago
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    it's not that simple lol..try it

  5. ganeshie8
    • 2 years ago
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    that is giving negative x. since these are two straight lines, there should be two values that satisfy the given condition right ?

  6. ganeshie8
    • 2 years ago
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    i mean two values of x

  7. shubhamsrg
    • 2 years ago
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    ok leme try 3x/7 = 4x/7 + 686 yes it gives -ve x it has no +ve solution i'd guess..

  8. Arthur_Bonnouvrier
    • 2 years ago
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    no positive values only -4802

  9. ganeshie8
    • 2 years ago
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    |dw:1340005766066:dw|

  10. shubhamsrg
    • 2 years ago
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    why 2 solns.. see 1st eqn gives a line passing through (0,0) and slope = 3/7 2nd eqn gives a line passing through (0,343) ,slope =2/7 and both lines are increasing.. its easy to conclude then,,why only -ve x will work..

  11. ganeshie8
    • 2 years ago
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    humm.. lets say x axis is time

  12. ganeshie8
    • 2 years ago
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    at t=0, y2 is 343 more than y1

  13. Arthur_Bonnouvrier
    • 2 years ago
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    just enter the equation y1=2*(y2) and you'll see that there is only one value which is -4802.

  14. ganeshie8
    • 2 years ago
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    since y1 is increasing much faster than y2, at some point y1 reaches y2

  15. ganeshie8
    • 2 years ago
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    after some more time, y1 value reaches double the y2 also right ?

  16. ganeshie8
    • 2 years ago
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    i need to know the time at which y1 reaches twice the value of y2. hope my question is making sense...

  17. Arthur_Bonnouvrier
    • 2 years ago
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    are you sure you entered the correct equations?

  18. ganeshie8
    • 2 years ago
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    there should be two values for x.. goes from \[-\infty \to + \infty\]

  19. ganeshie8
    • 2 years ago
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    i am going for lunch.. brb

  20. shubhamsrg
    • 2 years ago
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    well yes,,there'll be 2 values,,we didnt take the co-ordinates into account earlier,,setting y1=-2y2 will yield another solution,,but that also gives -ve x.. hmmn there must be some explanation,,i'll get back to it..

  21. ganeshie8
    • 2 years ago
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    sure there must be some explanation...

  22. shubhamsrg
    • 2 years ago
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    y1/y2 = 3x/(2x + 2401) =k(say) 3x = 2kx + 2401k =>x = 2401k/(3-2k) so for k=3/2,,x is undefined,,k > 3/2,,x gets -ve.. k<3/2,,x is +ve.. maybe this is satisfactory,,since 2>3/2,,x had to be -ve.. but why is x -ve for k = -2// hmm..

  23. shubhamsrg
    • 2 years ago
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    maybe k is ratio of magnitudes by default,, :|

  24. telliott99
    • 2 years ago
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    Much later.. I sketched the graph. It's clear that the two lines must cross. Since the relative slope (y1 grows faster than y2) is 1/7, that happens at 7 (343) = x = 2401. There, y1 = 3 (343) and y2 = 2 (343) + 343 = 1029. However, since y2 is moving away from the x-axis with a slope of 2/7, and y1 only grows faster than y2 by 1/7, I think there is no way y1 can ever be twice y2. Instead, at any x > 2401, (I think) y2 will be 2/3 of the way to y1 up from 1029. HTH.

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  25. ganeshie8
    • 2 years ago
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    i had trouble understanding this few days back... i guess i get it finally.. frm ur explanation i see : for (y1 = 3/7x) to reach twice the value of function (y2 = 2/7x+343) --- is same as --- asking for a slope of 4/7... but y1 is slow --- its growing at 3/7 only.... so y1 can never reach twice y2.... . !! thank you :)

  26. telliott99
    • 2 years ago
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    yw

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