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jagatuba
 5 years ago
Can anyone show me how to prove two equations are equivalent?
jagatuba
 5 years ago
Can anyone show me how to prove two equations are equivalent?

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jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0L*[c/(1 (1 + c) ^ n) ] and L*[c(1 + c)^n]/[(1 + c)^n  1]

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0Bot of those expressions should equal P

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0In other words: P = L*[c/(1 (1 + c) ^ n) ] and P = L*[c(1 + c)n]/[(1 + c)n  1]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0L , c, n are variables ???

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0Yes I get as far as eliminating L by division, but then I get stuck.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Could you use the equation button to express it in a better way....I m pretty gud at solving equations...but its just written weirdly...

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah I know what you mean. I'll try. I was messing with the equation button last night but was not getting the equations to look right. lol Give me a few

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0\[P=L \times \left[ c \div \left( 1\left( 1+c \right)^{n} \right) \right]\] \[P=L \times \left[ \left( c \times \left( 1+c \right)^{n} \right) \div \left( \left( 1+c \right)^{n} \right)1\right]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so simple ...it was...

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know if this will help you at all but these are two equations that are used to determine the monthly payment on a mortgage loan (P=monthly payment). I know they do come up with the same results I just have to prove mathematically

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in the first equation, make (1+c)^ n = 1/ (1+c)^ n and then take the lcm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes in the first equation...

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0I could write the right side as 1/(c+1)(c+1) but how would I write the left side with it's negative exponent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but that was (1+c)^n ...wasnt it?

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0Wait now I'm getting confused. lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK...let me make sure again....we are trying to prove that first equation equals to second equation...right??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in the first equation...take the lcm in the denominator

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0Okay give me a minute to work this out on paper and I'll post again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which will eventually go to numerator

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0SO after dividing both sides of the equation by L I get: \[\left( c \div \left( 1\left( 1+c \right)^{n} \right) \right)=\left( \left( c \times \left( 1+c \right)^{n} \right)\div \left( \left( 1+c \right)^{n} \right)1 \right)\] Next I take care of the negative exponent like this: \[c\left( 1+c \right)^{n }=\left( c \times \left( 1+c \right)^{n} \right)\div \left( \left( 1+c \right)^{n}1 \right)\] Is this right so far?

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0Actually that right side should be \[c \times \left( 1+c \right)^{n}\]

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0But now I'm lost again. Dang it.

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0But if I divide each side by \[\left( 1+c \right)^{n}\] I still end up with: \[c =c \div \left( \left( 1+c \right)^{n}1 \right)\]

jagatuba
 5 years ago
Best ResponseYou've already chosen the best response.0That can't be right can it?
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