Question from higher algebra - hall knight

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The discussion revolves around solving a higher algebra proof problem involving the manipulation of equations. The initial attempt rewrites the equation but leads to confusion regarding the provided solution. The key point of contention is understanding how the expressions were derived, particularly the relationship between the variables and constants. The participants express difficulty in grasping the transformation steps necessary to arrive at the final form of the answer. Clarification on these derivations is essential for a complete understanding of the proof.
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The Attempt at a Solution



basically i would rewrite the equation as,

(x/l)/(mb + nc - la) = (y/m)/(nc + la - mb) = (z/n)/(la + mb - nc)

upon adding,i get
(x/l + y/m + z/n)/(la+mb+nc)

but what they provided is,
(y/m + z/n)/2la = two similar expressions

==> (ny + mz)/a = (lz + nx)/b = (mx + ly)/c

i did not understand how did they get that.

since this is a proof problem i only used the "3. The Attempt at a Solution " section.
 

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Image is the question.
 
smart_worker said:
Image is the question.
As in:

attachment.php?attachmentid=68318&d=1396642300.jpg
 
SammyS said:
As in:

attachment.php?attachmentid=68318&d=1396642300.jpg

ya this is the question sammy.I tried to solve this but couldn't.The explanation they provided couldn't be understood too
 
two facts are being derived and then used
$$k=\frac{a}{b}=\frac{c}{d}=\frac{e}{f}\rightarrow k=\frac{c+e}{d+f}=\frac{e+a}{f+b}=\frac{a+c}{b+d}$$
and
$$k=\frac{c+b}{d}=\frac{a+c}{e}=\frac{b+a}{f}\rightarrow k=\frac{2a}{e+f-d}=\frac{2b}{d+f-e}=\frac{2c}{d+e-f}$$

So we take the given equation put it in the right form and transform it the first way then we put it in the form needed and transform again. finally we put it in the form of the answer.
 

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