Question from higher algebra - hall knight

  • Thread starter smart_worker
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    Algebra
In summary, the given equation can be rewritten as a series of similar expressions, which are then manipulated using two derived facts. The final result is a simplified form of the original equation. This process is used to solve the problem shown in the image.
  • #1
smart_worker
131
1

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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  • #2
Image is the question.
 
  • #3
smart_worker said:
Image is the question.
As in:

attachment.php?attachmentid=68318&d=1396642300.jpg
 
  • #4
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
 
  • #5
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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