Solving Homework Statement: Proving a/b=c/d=e/f to ace/bdf

  • Thread starter Thread starter Miike012
  • Start date Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 2K views
Miike012
Messages
1,009
Reaction score
0

Homework Statement



if a/b = c/d = e/f

then show that...

(a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3) = ace/bdf

Homework Equations




1.I understand how to prove it but I don't understand algebraically how they got from
a/b = c/d = e/f to... (a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3)
( or is this irrelevant and what I should focus on is the proving the theorem?)

2. second I don't understnad how (a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3) = ace/bdf?
the numerator does not simplify into the denominator UNLESS...
because a/b = c/d = e/f

(a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3) could also equal (k^4 + 2k^3 - 3k^4)/(k^4 + 2k^3 - 3k^4) = 1

and ace/bdf could equal k^3/k^3 = 1

therefore 1=1 is true?
 
Physics news on Phys.org
then
a=bp, c= dp, and e=fp


I understand that part... is there something that you are hinting at?
 
Miike012 said:
1.I understand how to prove it but I don't understand algebraically how they got from
a/b = c/d = e/f to... (a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3)
( or is this irrelevant and what I should focus on is the proving the theorem?)
It's irrelevant. The expression is a given. You don't care where it comes from.
2. second I don't understnad how (a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3) = ace/bdf?
the numerator does not simplify into the denominator UNLESS...
because a/b = c/d = e/f
The point of the problem is to show that if you assume a/b = c/d = e/f, then (a3b+2c2e-3ae2f)/(b4+2d2f-3bf3) = ace/bdf. All proofs are like this. You're given a set of assumptions, and you want to show that it then follows that some statement is true.
(a^3b + 2c^2e - 3ae^2f) / (b^4 + 2d^2f - 3bf^3) could also equal (k^4 + 2k^3 - 3k^4)/(k^4 + 2k^3 - 3k^4) = 1

and ace/bdf could equal k^3/k^3 = 1

therefore 1=1 is true?
It's certainly true if a, b, c, d, e, and f all equaled each other (and aren't equal to 0), the equality would hold, but that fact doesn't really help in your proof because you're using an assumption that wasn't a given in the problem. All you can assume is a/b = c/d = e/f.
 
wow, thank you ... that makes a lot more sense now.
 
A good way to write a proof like this is to start with one side of the equation and manipulate it until you end up with the other side:

[tex]\begin{align*}<br /> \frac{a^3b + 2c^2e - 3ae^2f}{b^4 + 2d^2f - 3bf^3} & = \cdots \\<br /> & = \cdots \\<br /> & \vdots \\<br /> & = \frac{abc}{def}<br /> \end{align*}[/tex]