Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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.

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

using geometry? maybe you mean using purely algebra?
ya solving equations... algebra is ok.. . :)
just set y1 = 2y2 make substitutions and solve for x

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

it's not that simple lol..try it
that is giving negative x. since these are two straight lines, there should be two values that satisfy the given condition right ?
i mean two values of x
ok leme try 3x/7 = 4x/7 + 686 yes it gives -ve x it has no +ve solution i'd guess..
no positive values only -4802
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..
humm.. lets say x axis is time
at t=0, y2 is 343 more than y1
just enter the equation y1=2*(y2) and you'll see that there is only one value which is -4802.
since y1 is increasing much faster than y2, at some point y1 reaches y2
after some more time, y1 value reaches double the y2 also right ?
i need to know the time at which y1 reaches twice the value of y2. hope my question is making sense...
are you sure you entered the correct equations?
there should be two values for x.. goes from \[-\infty \to + \infty\]
i am going for lunch.. brb
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..
sure there must be some explanation...
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..
maybe k is ratio of magnitudes by default,, :|
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.
1 Attachment
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 :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question