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Mathematics 7 Online
ganeshie8 (ganeshie8):

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.

OpenStudy (shubhamsrg):

using geometry? maybe you mean using purely algebra?

ganeshie8 (ganeshie8):

ya solving equations... algebra is ok.. . :)

OpenStudy (shubhamsrg):

just set y1 = 2y2 make substitutions and solve for x

OpenStudy (lgbasallote):

it's not that simple lol..try it

ganeshie8 (ganeshie8):

that is giving negative x. since these are two straight lines, there should be two values that satisfy the given condition right ?

ganeshie8 (ganeshie8):

i mean two values of x

OpenStudy (shubhamsrg):

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

OpenStudy (anonymous):

no positive values only -4802

ganeshie8 (ganeshie8):

|dw:1340005766066:dw|

OpenStudy (shubhamsrg):

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..

ganeshie8 (ganeshie8):

humm.. lets say x axis is time

ganeshie8 (ganeshie8):

at t=0, y2 is 343 more than y1

OpenStudy (anonymous):

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

ganeshie8 (ganeshie8):

since y1 is increasing much faster than y2, at some point y1 reaches y2

ganeshie8 (ganeshie8):

after some more time, y1 value reaches double the y2 also right ?

ganeshie8 (ganeshie8):

i need to know the time at which y1 reaches twice the value of y2. hope my question is making sense...

OpenStudy (anonymous):

are you sure you entered the correct equations?

ganeshie8 (ganeshie8):

there should be two values for x.. goes from \[-\infty \to + \infty\]

ganeshie8 (ganeshie8):

i am going for lunch.. brb

OpenStudy (shubhamsrg):

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..

ganeshie8 (ganeshie8):

sure there must be some explanation...

OpenStudy (shubhamsrg):

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..

OpenStudy (shubhamsrg):

maybe k is ratio of magnitudes by default,, :|

OpenStudy (anonymous):

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.

ganeshie8 (ganeshie8):

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 :)

OpenStudy (anonymous):

yw

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