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Mathematics 17 Online
OpenStudy (anonymous):

Divide: 1)2x+2/x^2+5x+6 / x^2+4x+3 2)x^2+2x-15/x-3 / x+5/6x-18

OpenStudy (anonymous):

\[1) \frac{ 2x+2 }{x^{2}+5x+6 } \div x^{2}+4x+3\]

OpenStudy (anonymous):

2)\[\frac{ x^{2}+2x-15 }{ x-3 } \div /\frac{x+5}{6x-18}\]

OpenStudy (anonymous):

@hartnn

hartnn (hartnn):

ok, try to factor whichever expression you can. go one by one 2x+2 =... ? (this is easy) x^2+5x+6 =... ?

OpenStudy (anonymous):

1) 1

OpenStudy (anonymous):

(x+3)(x+2)

hartnn (hartnn):

2x+2 = 1 ?? :O

hartnn (hartnn):

x^2+5x+6 = (x+3)(x+2) is correct :)

OpenStudy (anonymous):

Oh I thought it was 2x+2=0 haha

hartnn (hartnn):

2x+2 = 2 (x+1) got this ? x^2+4x+3 =... ?

OpenStudy (anonymous):

OH my mistake. So then the bottom would be ...x+2 x+1 .. No ..

hartnn (hartnn):

hmm .. ? ok, one by one 2x+2 = 2 (x+1) x^2+5x+6 = (x+3)(x+2) is correct. what about x^2+4x+3 =... ?

hartnn (hartnn):

we'll see tops and bottoms later... :P

OpenStudy (anonymous):

that would be .. (x+1)(x+3) Ah no , the last line I'm stuck on factoring o;

hartnn (hartnn):

yes, thats correct.

OpenStudy (anonymous):

Oh okay yay lol.

OpenStudy (anonymous):

\[\frac{ 2(x+1) }{(x+3)(x+2) } \div (x+1)(x+3)\]

hartnn (hartnn):

\(\large \dfrac{ 2(x+1) }{(x+3)(x+2) } \div (x+1)(x+3)=\dfrac{ 2(x+1) }{(x+3)(x+2) } \times \dfrac{1}{ (x+1)(x+3)}\) what gets cancelled ?

hartnn (hartnn):

also, you got that step ?

OpenStudy (anonymous):

2 / (x+2) = 1

hartnn (hartnn):

:O first just tell me what gets cancelled ?

OpenStudy (anonymous):

x+1, x+3

hartnn (hartnn):

sure, does 'x+3' really get cancelled ? and yes, x+1 gets cancelled.

OpenStudy (anonymous):

Oh no it doesn't. So then we're left with.. 2/x+3 x+2 = 1/x+3

hartnn (hartnn):

how is the numerator = 1 ?

OpenStudy (anonymous):

Oh no no , it's just what's on the other side.

OpenStudy (anonymous):

Oh I meant x

hartnn (hartnn):

\(\large \dfrac{ 2\cancel{(x+1)} }{(x+3)(x+2) } \times \dfrac{1}{ \cancel{(x+1)}( {x+3})}=\dfrac{2}{(x+2)(x+3)^2}\) is what remas and cannot be simplified, so the answer...

OpenStudy (anonymous):

Thank you so much again! I understand it so much better. One last problem? So sorry for wearing you out!

OpenStudy (anonymous):

The second equation now D: Haha . :) Then I'm done.

hartnn (hartnn):

no problem :) 6x-18 =... ?

hartnn (hartnn):

x^2+2x-15 =... ?

hartnn (hartnn):

umm...(x+6)(x-3) for what ? 6x-18 = 6 (......?.....)

OpenStudy (anonymous):

6(x+3)

hartnn (hartnn):

6 (x+3) or 6 (x-3) ??

OpenStudy (anonymous):

Ooh x-3 xD

hartnn (hartnn):

x^2+2x-15 =... ?

OpenStudy (anonymous):

Um.. I'm not sure? D:

OpenStudy (anonymous):

Oh wait.

OpenStudy (anonymous):

(x+5)(x-3)?

hartnn (hartnn):

yes!

OpenStudy (anonymous):

Yay.

hartnn (hartnn):

what gets cancelled ?

OpenStudy (anonymous):

x-3

OpenStudy (anonymous):

x+5

hartnn (hartnn):

yes :) what remains ?

OpenStudy (anonymous):

6(x-3) ?

OpenStudy (anonymous):

Oh wait. x-3 / 6(x-3)

hartnn (hartnn):

\(\large \frac{ x^{2}+2x-15 }{ x-3 } \div \frac{x+5}{6x-18} \\ = \large \frac{ x^{2}+2x-15 }{ x-3 } \times \frac{6x-18}{x+5} = \dfrac{\cancel{(x+5)}\cancel{(x-3)}}{\cancel{x-3}} \dfrac{6(x-3)}{\cancel{x+5}}\) what remains ?

OpenStudy (anonymous):

Just 6(x-3) ? Like i originally thought lol. That's the final answer?

hartnn (hartnn):

yes! :P

OpenStudy (anonymous):

Thank you so very much @hartnn ! :) Have a good night xo

OpenStudy (anonymous):

I appreciate it!

hartnn (hartnn):

well, you have an awesome night , and sweet dreams ^_^ its morning here, i'll start my day :)

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