## Jamierox4ev3r one year ago More math review 5c. Simplify the rational expression

1. Jamierox4ev3r

Give me a sec while I type this out

2. Jamierox4ev3r

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

3. Jamierox4ev3r

^and that's the thing I need to simplify. Thanks for your patience

4. Zale101

Whenever i have complex expressions like this, i just start off by using this rule. It applies to all fractions added or subtracted together. $$\LARGE \frac{a}{b}-\frac{c}{d}=\frac{a*d-c*d}{b*d}$$

5. Jamierox4ev3r

wait, you would do this so you can have a common denominator right? :o

6. Zale101

$$\LARGE \frac{x^2}{x^2-4}-\frac{x+1}{x+2}=\frac{x^2*(x+2)-(x^2-4)*(x+1)}{(x^2-4)(x+2)}$$

7. Zale101

Yes, jamie. That's the shortest cut to get the common denominator.

8. Zale101

In my opinion =)

9. Nnesha

*or* first you can factor out the x^2 -4 :=)) and then find common denominator

10. Zale101

I'll have to go now, Nnesha will take over :)

11. Nnesha

nah jami can solve it! :P

12. Jamierox4ev3r

ah I see. Makes sense. Thank you thus far @Zale101 ^_^ and @Nnesha I will certainly try my best ^O^ but even if i don't factor out the $$x^{2}-4$$ beforehand, I would factor it out when simplifying the equation hmm so let's see....

13. Nnesha

ye

14. Jamierox4ev3r

So that would give me $$\Large\frac{x^{2}(x+2)-(x-2)(x+2)(x+1)}{(x-2)(x+2)(x+2)}$$

15. Jamierox4ev3r

Am I on the right track?

16. Nnesha

yes right

17. Jamierox4ev3r

cool. Hold on though...the subtraction sign is throwing me in for a bit of a loop. would that just mean that (x+2) and (x-2) in the numerator cancel each other out?

18. Nnesha

do you mean theses two parenthess http://prntscr.com/88446s

19. Nnesha

these*

20. Jamierox4ev3r

yes

21. Nnesha

let's say x+2 = y x+1=b x-2=z then can you combine these?are there any like terms ?$\huge\rm x^2y - zyb$

22. Jamierox4ev3r

I think there are like terms..but that's kind of confusing. Can you maybe clarify? Basically, I've already expanded everything, but I'm just a little confused on the next steps I should take

23. Nnesha

yeah i'm sure you are familiar with *like terms * 2x+3x you can combine these two terms (2+3)x

24. Nnesha

here are some example of like terms 3x^2+4x^2 same variable with same xponent you can combine them but you can't combine $x^2+x$

25. Jamierox4ev3r

oh wait i see what you did, you equated the values in parentheses to variables. whoops i was wondering, "where did the y, z, and b come from? O_O" but I see. and yes I am familiar with combining like terms. There are like terms in the numerator, I'm just not sure which ones I can combine and how

26. Nnesha

ye we supposed that x+2 represent z

27. Nnesha

and no there aren't any like terms in the numerator

28. Jamierox4ev3r

oh right, because there is x^2. So what do you do?

29. Nnesha

you wouldn't subtract it bec they are not like terms but you can divide $$\huge\rm \frac{x^{2}\color{Red}{(x+2)}-(x-2)(x+2)(x+1)}{(x-2)\color{Red}{(x+2)}(x+2)}$$

30. Jamierox4ev3r

oh right! and those would cancel out, wouldn't they?

31. Jamierox4ev3r

or at the very least, be equivalent to 1.

32. Nnesha

you wouldn't subtract it bec they are not like terms but you can divide $$\huge\rm \frac{x^{2}\color{Red}{(x+2)}-(x-2)\color{Red}{(x+2)}(x+1)}{(x-2)\color{Red}{(x+2)}(x+2)}$$ the other x+2 is also dividing by x+2

33. Jamierox4ev3r

$$\huge\rm \frac{x^{2}\color{Red}{(x+2)}-(x-2)\color{Red}{(x+2)}(x+1)}{(x-2)\color{Red}{(x+2)}\color{green}{(x+2)}}$$ what about the one in green? could you also divide by that?

34. Nnesha

no you divide it by only one like $\frac{ \cancel{x} }{ x \times \cancel{x} }$

35. Nnesha

i have to go so urgent

36. Nnesha

sorry will come back later to check ur work :(

37. Jamierox4ev3r

okay then. So if you divide what's in red, then you would have $$\huge\frac{x^{2}-(x-2)(x+1)}{(x-2)(x+2)}$$ right? and from here, it looks like you could also divide these: $$\huge\frac{x^{2}-\color{red}{(x-2)}(x+1)}{\color{red}{(x-2)}(x+2)}$$ Is that a correct assumption? awwwee come back soon!

38. Plasmataco

Looks right to me

39. laughoutloud

I think there was an easier way to solve this

40. laughoutloud

@Jamierox4ev3r

41. Jamierox4ev3r

How?

42. laughoutloud

If you factor out the x^2-4 the common denominator is (x-2)(x+2)

43. Jamierox4ev3r

the x^2-4 was factored, just a little later in the process.

44. laughoutloud

$\frac{ x^{2} }{ (x-2)(x+2) }-\frac{ (x+1)(x-2)}{ (x-2)(x+2) }$

45. laughoutloud

then when I subtract I get $\frac{ x+2 }{ (x-2)(x+2) }$

46. laughoutloud

your final answer is $\frac{ 1 }{ x-2 }$

47. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Jamierox4ev3r okay then. So if you divide what's in red, then you would have $$\huge\frac{x^{2}-(x-2)(x+1)}{(x-2)(x+2)}$$ right? and from here, it looks like you could also divide these: $$\huge\frac{x^{2}-\color{red}{(x-2)}(x+1)}{\color{red}{(x-2)}(x+2)}$$ Is that a correct assumption? awwwee come back soon! $$\color{blue}{\text{End of Quote}}$$ yep that's right $\frac{ x^2-(x+1) }{ x+2}$ now tere is negative sign at the numerator so you should distribute x+1 by -1

48. Jamierox4ev3r

I see how you would get the final answer from the previous, but I don't get how you subtracted. (Oh, and for future reference, never provide answers. I would have liked to figure it out myself. But other than that, you were doing a great job of helping me. Thank you ^^)

49. laughoutloud

The subtraction sign is distributed to both the (x+1) and the (x-2)

50. laughoutloud

try this https://www.mathway.com/

51. Nnesha

ye right there is a typo

52. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Nnesha *or* first you can factor out the x^2 -4 :=)) and then find common denominator $$\color{blue}{\text{End of Quote}}$$ like i said earlier yes you can find factor of x^2-4 first and then find common denominator it will it easier to understand and that's what my teacher taught us

53. Jamierox4ev3r

@Nnesha if I use this method, then I would have $$\huge\frac{x^{2}-x-1}{x+2}$$ Still not sure what I would do from there though. and @laughoutloud distributing the negative would give you -x-1-x+2, so I'm still struggling to see how you got from the original to the subtraction.

54. Nnesha

you should foil (x-2)(x+1) and then distribute by negative sign it will give you LIKE terms to combine

55. laughoutloud

yes sorry that is what i did

56. Jamierox4ev3r

Alright. I think I'm starting to get it. Just lemme figure this out on my own, I have to go in a bit. But thank you guys for all your help. I definitely have a better grasp on how to approach these types of problems.

57. Jamierox4ev3r

I'll be back later, and I'll open up a new question. But @laughoutloud one last question for ya. $$\Huge\frac{ x^{2} }{ (x-2)(x+2) }-\frac{ (x+1)\color{red}{(x-2)}}{\color{red}{ (x-2)}(x+2) }$$ So for the things marked in red...how did you get that?

58. laughoutloud

I had to make the denominators the same so I multiplied (x-2) to the numerator and denominator

59. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Jamierox4ev3r okay then. So if you divide what's in red, then you would have $$\huge\frac{x^{2}-(x-2)(x+1)}{(x-2)(x+2)}$$ right? and from here, it looks like you could also divide these: $$\huge\frac{x^{2}-\color{red}{(x-2)}(x+1)}{\color{red}{(x-2)}(x+2)}$$ Is that a correct assumption? awwwee come back soon! $$\color{blue}{\text{End of Quote}}$$ here no you can't divide x-2 bec that's not a common denominator of both fraction

60. Jamierox4ev3r

Oh I see :o wao, solving that way is much confusing XD alright, I'll open up a new question. Teach me the method that your teacher taught you, alright? :P

61. Nnesha

i'll suggest to factor out the expression first and then find common denominator like laughoutlout mentioned

62. Nnesha

sure

63. Jamierox4ev3r

right. you mentioned that. I'd just like to open up a new post since this one is getting laggy for me. ty! ^_^