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Mathematics 16 Online
OpenStudy (anonymous):

What is the simplified form of (x+8/3x+7) + (x+7/x+4)?

OpenStudy (anonymous):

I'm just confused about whether or not I should foil anything or not

OpenStudy (cathyangs):

Assuming the / are divisions, then yes you need to do some foiling. (x+8)/(3x+7) + (x+7)/(x+4) you multiply to have a common denominator. Does this make sense?

OpenStudy (anonymous):

kinda? I mean would I have to multiply the second fraction by 3?

OpenStudy (anonymous):

so then you'd have if it's 3x+21/3x+12, what would be the next step? just foil it?

OpenStudy (cathyangs):

No, you should multiply the first fraction by (x+4)/(x+4) and the second fraction by (3x+7)/(3x+7). Those are both equivalent to multiplying by 1. Then foil. Does this make sense?

OpenStudy (anonymous):

can you show me what that would look like though

OpenStudy (anonymous):

i mean in the equation thing

OpenStudy (cathyangs):

So you want the two fractions to have a common denominator, which will be (3x+7)(x+4) on the bottom, right? \[\frac{ x+8 }{ 3x+7 } * \frac{ x+4 }{x+4} \] for the first one, is the same as: \[\frac{ (x+8)(x+4) }{ (3x+7)(x+4) } = \frac{ x^{2} + 12x + 32}{ 3x ^{2} + 19x + 28}\] I just did FOIL using the top row (and added the bx terms together in my head). Does this make sense?

OpenStudy (anonymous):

OHH ok yea. My only thing is that in my answer bank, the denominator is correct but the numerator isn't. So basically: \[4x^2+40x+81/3x^2+19x+28\]

OpenStudy (anonymous):

@cathyangs did u leave :/

OpenStudy (anonymous):

http://www.cymath.com/answer.php?q=(x%2B8%2F3x%2B7)%20%2B%20(x%2B7%2Fx%2B4) This Website should help :)

OpenStudy (cathyangs):

Sorry, I had to work out my own answer to the problem. You got it right! :)

OpenStudy (anonymous):

ok thank u!!

OpenStudy (anonymous):

Acutally, before I leave, I was wondering if you could just confirm if I'm right about this one or not?: \[\frac{ x+8 }{ 4 } - \frac{ x+5 }{ 4 }\] I got \[\frac{ 3 }{ 4}\]

OpenStudy (anonymous):

You're correct!!

OpenStudy (anonymous):

thanks

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