18jonea
  • 18jonea
find the domain and simplify
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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18jonea
  • 18jonea
\[\frac{ x+1 }{ 3x-3 } - \frac{ 4 }{ x^2-3x+2 }\]
Michele_Laino
  • Michele_Laino
we have to factorize both denominators
18jonea
  • 18jonea
i know and i did the first but dont know how to do the decond first one is 3(x-1)

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Michele_Laino
  • Michele_Laino
for the second one, you have to search for two numbers such that their sum is 3, and their product is 2
18jonea
  • 18jonea
1 and 2
Michele_Laino
  • Michele_Laino
correct!
Michele_Laino
  • Michele_Laino
so we can write this: \[\Large {x^2} - 3x + 2 = \left( {x - 1} \right)\left( {x - 2} \right)\]
18jonea
  • 18jonea
now we have to find the domain
Michele_Laino
  • Michele_Laino
therefore: \[\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
yes! we have to find the domain
Michele_Laino
  • Michele_Laino
in order to do that, we note that we can not divide by zero
18jonea
  • 18jonea
Domain is all real numbers except when x = 1 or 2
18jonea
  • 18jonea
is that right?
Michele_Laino
  • Michele_Laino
so we have to request that: \[\Large \begin{gathered} x - 1 \ne 0 \hfill \\ \left( {x - 1} \right)\left( {x - 2} \right) \ne 0 \hfill \\ \end{gathered} \] yes! correct!
18jonea
  • 18jonea
ok now we have to find the common denominator
Michele_Laino
  • Michele_Laino
right! we have to find the common denominator between the two denominators: \[\Large \begin{gathered} 3\left( {x - 1} \right) \hfill \\ \left( {x - 1} \right)\left( {x - 2} \right) \hfill \\ \end{gathered} \]
18jonea
  • 18jonea
3?
Michele_Laino
  • Michele_Laino
no, we have to take all uncommon and common factors with the highest exponent
18jonea
  • 18jonea
ok so how do we do that
Michele_Laino
  • Michele_Laino
it is simple, x-1 is a common factor, whereas 3 and x-2 are uncommon factor, so the common denominator is: 3*(x-1)*(x-2)
Michele_Laino
  • Michele_Laino
and hence, we can write this: \[\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{...}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered} \]
18jonea
  • 18jonea
so what is the common denominator
Michele_Laino
  • Michele_Laino
it is: \[\Large {3\left( {x - 1} \right)\left( {x - 2} \right)}\]
Michele_Laino
  • Michele_Laino
now we have to divide the common denominator by the first numerator, namely what is: \[\Large \frac{{3\left( {x - 1} \right)\left( {x - 2} \right)}}{{3\left( {x - 1} \right)}} = ...?\]
Michele_Laino
  • Michele_Laino
you can cancel out common factors
Michele_Laino
  • Michele_Laino
|dw:1440013189699:dw|
18jonea
  • 18jonea
why are you cancelling
Michele_Laino
  • Michele_Laino
because one factor is at denominator, and the same factor is at denominator
Michele_Laino
  • Michele_Laino
please wait a moment
Michele_Laino
  • Michele_Laino
the result is x-2, so the next step is: \[\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x + 1} \right)\left( {x - 2} \right) - 4 \cdot ....}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
now we have to divide the common denominator by the second denominator, namely: \[\Large \frac{{3\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = ...\]
18jonea
  • 18jonea
i dont think this is right
Michele_Laino
  • Michele_Laino
yes! it is right :)
18jonea
  • 18jonea
you have to do the common denominator for each part of the fraction
Michele_Laino
  • Michele_Laino
the common denominator is only one
18jonea
  • 18jonea
so for the first part wouldnt it be \[\frac{ x+1(x-2) }{ 3(x-1)(x-2)}\]
Michele_Laino
  • Michele_Laino
that's right!
Michele_Laino
  • Michele_Laino
now, we have this: \[\Large \frac{{3\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = 3\]
Michele_Laino
  • Michele_Laino
so we can write this: \[\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x + 1} \right)\left( {x - 2} \right) - 4 \cdot 3}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
we have to simplify the numerator: \[\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x + 1} \right)\left( {x - 2} \right) - 4 \cdot 3}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{{x^2} - x - 2 - 12}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} = \frac{{{x^2} - x - 14}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered} \]
18jonea
  • 18jonea
and then that would be the final answer?
Michele_Laino
  • Michele_Laino
yes!
18jonea
  • 18jonea
Thank you
Michele_Laino
  • Michele_Laino
:)

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