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I encountered this math question on an act test, and I just cannot come up with the answer they have.

(200)/(r+10)= (200)/(r)-1

By Benzinspeicher (Benzinspeicher) on Friday, August 06, 2004 - 12:23 am: Edit

first, change the right side of the equation to make it all have one common denominator: I see that it's (200/r) - 1, so the 1 is the same as (r/r), so the common denominator is "r", so the right side of the equation is (200-r)/(r); next i write out the whole thing: (200)/(r+10)=(200-r)/(r); Next, I cross multiply, and get r^2+10r-2000=0. So I factor it out and get (r+50)(r-40)=0
So r+50=0, and r-40=0. Solve them, and I get r=40, and r=-50

Hope this helps

By Tongos (Tongos) on Friday, August 06, 2004 - 12:50 am: Edit

should come out with 40 or -5o

apply
200/(r+10)=(200-r)/r=(200r=(200-r)(r+10)) = (-r^2-10r+2000) and solve.

an alternate method would be an iteration formula for the equation:
give the two equations as follows
200/(r+10)-(200)/(y)=-1
sub a value in for r, lets say r=30, so y=33.3333333 (pick values for which y-r is small, but a rough approx is good)
do the derivative of the first part: 200/(r+10)
-200/((r+10)^2) multiply it by r. and add this to the (derivative of the second part multiplied by y) (200y/y^2)
we should get this = -200r/((r+10)^2)+200/y
now we divide by the derivatives added together.

{-200r/((r+10)^2)+200y/y^2}/{(-200)/((r+10)^2)+(200/y^2)} plug the values that you subbed for r and y back into the equation to get a very close answer (40.909090909). If your not satisfied, do it again, make 40.909090909 r. and do it again until you come out with an exact answer.
This isnt probably the easiest method for an equation like this (probably really dumb too, (cause its easy to solve by algebra) but if your given any kind of equation where you have it implicitly defined, use this. Or if your'e given a polynomial and asked to find the zeros of it, use this.
Trust me, the method converges down to the answer very quickly.

By Tongos (Tongos) on Friday, August 06, 2004 - 01:18 am: Edit

i really hoped that helped, it can also solve stuff like 10x=200, fred is having trouble solving this. he wants to use a formula.
break it up!
9X+Y=200
the formula says worry about the coeffs
he plugs in 1 for "x". x=1 and y=191.
uses the rule {(9)(1)+(191)(1)}/{9+1) does the math (he's knows arithmetic) and he comes out with the answer!

By Benzinspeicher (Benzinspeicher) on Friday, August 06, 2004 - 01:19 am: Edit

WHY DO PEOPLE USE CALCULUS TO SOLVE ALGEBRA PROBLEMS?????????

Maybe it's just cuz I hate certain things about calculus: not much logic involved-more knowledge than anything else.

By Tongos (Tongos) on Friday, August 06, 2004 - 11:17 am: Edit

actually benzinspeicher, while it does cover knowledge, the theory did require me to think about normal logic things, like weighted averages, rate of change, geometry, several variable equations, defining functions and not just knowing calculus (but understanding it).
Anyway, i thought the theory is really cool, because it has the ability to solve equations that you would normally use a matrix on. It can solve just about any equation by covergence.