simplifying complex fractions: (a/a-1 - 2a/a-2)/(2a/a-2)-(3a
- Thread starter iamthere
- Start date
I'll step thorugh the first one. You do the other.
\(\displaystyle \L\\\frac{\frac{a}{a-1}-\frac{2a}{a-2}}{\frac{2a}{a-2}-\frac{3a}{a-3}}\)
You could just cross-multiply the top and bottom:
\(\displaystyle \L\\\frac{\frac{a(a-2)-2a(a-1)}{(a-1)(a-2)}}{\frac{2a(a-3)-3a(a-2)}{(a-2)(a-3)}}\)
Distribute:
\(\displaystyle \L\\\frac{\frac{-a^{2}}{(a-1)(a-2)}}{\frac{-a^{2}}{(a-2)(a-3)}}\)
It's the same as:
\(\displaystyle \L\\\frac{-a^{2}}{(a-1)(a-2)}\cdot\frac{(a-2)(a-3)}{-a^{2}}\)
There. Now you have several hours to finish the other.
\(\displaystyle \L\\\frac{\sout{(a-2)}(a-3)}{(a-1)\sout{(a-2)}}\)
\(\displaystyle \L\\\frac{\frac{a}{a-1}-\frac{2a}{a-2}}{\frac{2a}{a-2}-\frac{3a}{a-3}}\)
You could just cross-multiply the top and bottom:
\(\displaystyle \L\\\frac{\frac{a(a-2)-2a(a-1)}{(a-1)(a-2)}}{\frac{2a(a-3)-3a(a-2)}{(a-2)(a-3)}}\)
Distribute:
\(\displaystyle \L\\\frac{\frac{-a^{2}}{(a-1)(a-2)}}{\frac{-a^{2}}{(a-2)(a-3)}}\)
It's the same as:
\(\displaystyle \L\\\frac{-a^{2}}{(a-1)(a-2)}\cdot\frac{(a-2)(a-3)}{-a^{2}}\)
There. Now you have several hours to finish the other.
\(\displaystyle \L\\\frac{\sout{(a-2)}(a-3)}{(a-1)\sout{(a-2)}}\)
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- Apr 12, 2005
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Re: Help! Complex fractions
First, one must understand the problem statement.
a/a-1 = (a/a)-1 = 1-1 = 0 Is that what you mean? I suspect not.
a/(a-1) is a different animal.
You should find common denominators and perform the indicated operations.
There is lots of help, but this is not a panic site.iamthere said:
First, one must understand the problem statement.
a/a-1 = (a/a)-1 = 1-1 = 0 Is that what you mean? I suspect not.
a/(a-1) is a different animal.
You should find common denominators and perform the indicated operations.
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- Feb 4, 2004
- Messages
- 16,582
Your formatting is confusingly ambiguous. Did the one tutor guess your meaning correctly for the first exercise? Do you mean the following for the second one?iamthere said:
. . . . .\(\displaystyle \L \frac{\left(\frac{a\, -\, 2}{a\, -\, 1}\right)}{\left(\frac{a\, +\, 1}{a\, -\, 1}\, +\, (a\, +\, 1)\right)}\)
Note: This can also be written, in straight text, as:
. . . . .[(a - 2) / (a - 1)] / [(a + 1) / (a - 1) + (a + 1)]
Please reply with confirmation or correction. When you reply, please show what you have tried and how far you have gotten.
Thank you.
Eliz.