## 18jonea one year ago find the domain and simplify

1. 18jonea

$\frac{ x+1 }{ 3x-3 } - \frac{ 4 }{ x^2-3x+2 }$

2. Michele_Laino

we have to factorize both denominators

3. 18jonea

i know and i did the first but dont know how to do the decond first one is 3(x-1)

4. Michele_Laino

for the second one, you have to search for two numbers such that their sum is 3, and their product is 2

5. 18jonea

1 and 2

6. Michele_Laino

correct!

7. Michele_Laino

so we can write this: $\Large {x^2} - 3x + 2 = \left( {x - 1} \right)\left( {x - 2} \right)$

8. 18jonea

now we have to find the domain

9. Michele_Laino

therefore: $\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered}$

10. Michele_Laino

yes! we have to find the domain

11. Michele_Laino

in order to do that, we note that we can not divide by zero

12. 18jonea

Domain is all real numbers except when x = 1 or 2

13. 18jonea

is that right?

14. 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!

15. 18jonea

ok now we have to find the common denominator

16. Michele_Laino

right! we have to find the common denominator between the two denominators: $\Large \begin{gathered} 3\left( {x - 1} \right) \hfill \\ \left( {x - 1} \right)\left( {x - 2} \right) \hfill \\ \end{gathered}$

17. 18jonea

3?

18. Michele_Laino

no, we have to take all uncommon and common factors with the highest exponent

19. 18jonea

ok so how do we do that

20. 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)

21. Michele_Laino

and hence, we can write this: $\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{...}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered}$

22. 18jonea

so what is the common denominator

23. Michele_Laino

it is: $\Large {3\left( {x - 1} \right)\left( {x - 2} \right)}$

24. Michele_Laino

now we have to divide the common denominator by the first numerator, namely what is: $\Large \frac{{3\left( {x - 1} \right)\left( {x - 2} \right)}}{{3\left( {x - 1} \right)}} = ...?$

25. Michele_Laino

you can cancel out common factors

26. Michele_Laino

|dw:1440013189699:dw|

27. 18jonea

why are you cancelling

28. Michele_Laino

because one factor is at denominator, and the same factor is at denominator

29. Michele_Laino

30. Michele_Laino

the result is x-2, so the next step is: $\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x + 1} \right)\left( {x - 2} \right) - 4 \cdot ....}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered}$

31. Michele_Laino

now we have to divide the common denominator by the second denominator, namely: $\Large \frac{{3\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = ...$

32. 18jonea

i dont think this is right

33. Michele_Laino

yes! it is right :)

34. 18jonea

you have to do the common denominator for each part of the fraction

35. Michele_Laino

the common denominator is only one

36. 18jonea

so for the first part wouldnt it be $\frac{ x+1(x-2) }{ 3(x-1)(x-2)}$

37. Michele_Laino

that's right!

38. Michele_Laino

now, we have this: $\Large \frac{{3\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = 3$

39. Michele_Laino

so we can write this: $\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x + 1} \right)\left( {x - 2} \right) - 4 \cdot 3}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered}$

40. Michele_Laino

we have to simplify the numerator: $\Large \begin{gathered} \frac{{x + 1}}{{3x - 3}} - \frac{4}{{{x^2} - 3x + 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{3\left( {x - 1} \right)}} - \frac{4}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x + 1} \right)\left( {x - 2} \right) - 4 \cdot 3}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} = \hfill \\ \hfill \\ = \frac{{{x^2} - x - 2 - 12}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} = \frac{{{x^2} - x - 14}}{{3\left( {x - 1} \right)\left( {x - 2} \right)}} \hfill \\ \end{gathered}$

41. 18jonea

and then that would be the final answer?

42. Michele_Laino

yes!

43. 18jonea

Thank you

44. Michele_Laino

:)