need help simplifying complex fractions.... :( please help... i made an attempt.. dont just post the answer please teach me how to....

- anonymous

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- anonymous

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- anonymous

it's all about making the the bottom fraction 1... just a sec

- anonymous

ya i get that but i confuse myself... i know you need a common demnominator..

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- anonymous

ugh i keep messing uppppp

- anonymous

for yours, you have to get simple fraction over simple fraction first...

- anonymous

\[\frac{ 2-\frac{ 4 }{ x+2 } }{5+\frac{ 10 }{ x+2 } }\] is what you have but you need a simple fraction over a simple fraction. to get this, we start with a common denominator for the top expression...
\[2-\frac{ 4 }{ x+2 }=2\left( \frac{ x+2 }{ x+2 } \right)-\frac{ 4 }{ x+2 }=\frac{ 2x+4 }{ x+2 }-\frac{ 4 }{ x+2 }=\frac{ 2x }{ x+2 }\]
does this make sense?

- anonymous

yes that makes sense

- anonymous

so you do the same to the bottom fraction?

- anonymous

so now we do the same with the bottom expression. Would you like to try?

- anonymous

yes give me a min to do it

- anonymous

okay

- anonymous

|dw:1440524508903:dw|

- anonymous

excuse the terrible penmanship lol

- anonymous

very good!!!

- anonymous

so now we have
\[\Large{\frac{ \frac{ 2x }{ x+2 } }{ \frac{ 5x+20 }{ x+2 } }}\] right?

- anonymous

|dw:1440524931675:dw|

- anonymous

and then i do this?

- anonymous

yes, do you know why?

- anonymous

because its like dividing regular fractions... you flip and multiply... sooo the (x+2) crosses out right?

- anonymous

well, the idea is to make an equivalent expression that is a simple fraction rather than a complex fraction. This is because we humans have a much easier time understanding what a simple fraction represents compared to a complex fraction.
to make a simple fraction from a complex fraction, we simply manipulate the bottom fraction so that it is 1. we do this by multplying by the reciprocal of the bottom fraction, like this...
\[\Large{\frac{ \frac{ 2x }{ x+2 } }{ \frac{ 5x+20 }{x+2 } }=\frac{ \frac{ 2x }{ x+2 }\cdot \left( \frac{ x+2 }{ 5x+20 } \right) }{ \frac{ 5x+20 }{x+2 }\cdot \left( \frac{ x+2 }{ 5x+20 } \right)}=\frac{ \frac{ 2x }{ x+2 }\cdot \left( \frac{ x+2 }{ 5x+20 } \right) }{ \normalsize{1}}}\]

- anonymous

anything over 1 is just itself so that's why we end up with \[\frac{ 2x }{ x+2 }\cdot \frac{ x+2 }{ 5x+20 }\]
and we simplify from there

- anonymous

does that make sense?

- anonymous

yes so you factor 5x+20... to get 5(x+4)

- anonymous

\[\frac{ 2x }{ 5(x+4)}\] is the answer

- anonymous

or should i not factor it?

- anonymous

oh waiiittt hold on

- anonymous

wait nvm thats my final answer lol

- anonymous

to factor or not to factor is a matter of preference... you still have reduced to lowest terms.

- anonymous

but does the process make sense to you?

- anonymous

yeah it makes sense to focus on the numerator and denominator separately before you put it all together.. i was trying to do it all in one stepp. so thats my final correct answer?

- anonymous

yep

- anonymous

good job!

- anonymous

thank you for actually teaching me.. no one else takes the time to do that
i have 5 more to do lol. but would you be willing to check one of them for me after i do it.

- anonymous

sure

- anonymous

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- anonymous

and i think that can be reduced?

- anonymous

correct, it can be reduced

- anonymous

well this is a silly brain fart but i would reduce all the terms by 2?

- anonymous

\[\frac{ 4x-12 }{ 5x}\]

- anonymous

yep, that's the extent of it

- anonymous

one more question lol

- anonymous

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- anonymous

for the first question the denominator isnt a fraction...

- anonymous

you can make it one...
\[5+a=\frac{ 5+a }{ 1 }\]

- anonymous

ohhhh wait i got thiis

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- anonymous

yes except it is \(25-a^2 \text{ not } a^2 - 25\)

- anonymous

gotchaaa.. but does the order i write it affect it? i thought it was a difference of squares either way

- anonymous

true, but \(25-a^2 \ne a^2-25\) unless \(a=\pm 5\)

- anonymous

ahh i gotchaa. seriously thank you so much! youre amazingg! i might message you when im in need of a lesson lol

- anonymous

right on! good luck, have a great day and keep on learning!

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