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Mathematics 11 Online
OpenStudy (anonymous):

*** I will give you a medal if you help/answer *** Simplify the sum. State any restrictions on the variables. (x-2)/(x+3) + (10x)/(x^2-9)

OpenStudy (anonymous):

@tkhunny

OpenStudy (tkhunny):

It should be realatively obvious after you factor the right-side denominator.

OpenStudy (anonymous):

do I switch it to multiplication and the recep after i factor it

OpenStudy (tkhunny):

?? Y ou first look at it and decide what will make the denominator zero. This will be the restrictions. After that, just add like any other fractions.

OpenStudy (anonymous):

i don't know how... okay i factored.. and i got (x+3)(x-3) and that cancels out the (x+3)'s .. now what do i do ?

OpenStudy (anonymous):

I learn step by step... soo could you go through it with me?

OpenStudy (tkhunny):

Here's a plan. Never use he word "cancel" again, unless you are talking about your CableTV Service. \(\dfrac{x-2}{x+3} + \dfrac{10x}{x^{2} - 9} = \dfrac{x-2}{x+3} + \dfrac{10x}{(x+3)(x-3)}\)

OpenStudy (anonymous):

okay.

OpenStudy (tkhunny):

You need to remember how to add fractions. Common denominator? \(\dfrac{x-2}{x+3} + \dfrac{10x}{x^{2} - 9} = \dfrac{x-2}{x+3} + \dfrac{10x}{(x+3)(x-3)}= \dfrac{x-2}{x+3}\cdot\dfrac{x-3}{x-3} + \dfrac{10x}{(x+3)(x-3)}\) That is required to look VERY familiar.

OpenStudy (anonymous):

okayy now what ???

OpenStudy (tkhunny):

We have common denominator. What shall we do next? Who told you that you learn "step-by-step"? That's kind of silly. Thinking that prevents you from learning other ways. Think back a few years. How did you add \(\dfrac{1}{3} + \dfrac{5}{6}\)?

OpenStudy (anonymous):

I don't remember.. I honestly don't which is why I got on here to get help because my teachers teach me a pelletty as way and it's something I just can't follow. I'm sorry :(

OpenStudy (anonymous):

pelletty way *

OpenStudy (anonymous):

what the heck.. it wont let me say s h i t t y

OpenStudy (tkhunny):

Good. If you start talking that way you have to stop talking to me. If you cannot add 1/3 and 5/6, you cannot do this problem. No offense intended, it's just the way it is.

OpenStudy (anonymous):

cool. glad your going to help me.. isn't that what this site is for?

OpenStudy (tkhunny):

Show me that you can add 1/3 ad 5/6 and we can talk. Some of the help you need seems to be a little reality therapy. I am quite happy to help you. I am not happy to waste your time if you do not have the necessary prerequisites to advance through the material.

OpenStudy (anonymous):

obviously @jim_thompson5910 i need you to help me .

OpenStudy (tkhunny):

Seriously, just add 1/3 and 5/6. It is not much to ask.

OpenStudy (anonymous):

@jim_thompson5910

OpenStudy (anonymous):

i don't know how !!!!

jimthompson5910 (jim_thompson5910):

you already found that (x-2)/(x+3) + (10x)/(x^2-9) (x-2)/(x+3) + (10x)/((x+3)(x-3)) So the LCD is (x+3)(x-3)

jimthompson5910 (jim_thompson5910):

you need that first fraction to have the same denominator in order to add the two fractions

jimthompson5910 (jim_thompson5910):

how do we do this

OpenStudy (anonymous):

okayy. you multiply everything by the LCD to cancel out the fractions.

jimthompson5910 (jim_thompson5910):

well you can only use that trick when you have an equals sign

jimthompson5910 (jim_thompson5910):

but we don't have that here, so we just have to add the fractions like normal

OpenStudy (anonymous):

ohh okay.

jimthompson5910 (jim_thompson5910):

what is the first denominator missing to get it equal to the LCD

OpenStudy (anonymous):

the (x-3)

jimthompson5910 (jim_thompson5910):

so you need to multiply top and bottom of the first fraction by (x-3)

jimthompson5910 (jim_thompson5910):

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