anonymous
  • anonymous
need help simplifying complex fractions.... :( please help... i made an attempt.. dont just post the answer please teach me how to....
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
it's all about making the the bottom fraction 1... just a sec
anonymous
  • anonymous
ya i get that but i confuse myself... i know you need a common demnominator..

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anonymous
  • anonymous
ugh i keep messing uppppp
anonymous
  • anonymous
for yours, you have to get simple fraction over simple fraction first...
anonymous
  • 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
  • anonymous
yes that makes sense
anonymous
  • anonymous
so you do the same to the bottom fraction?
anonymous
  • anonymous
so now we do the same with the bottom expression. Would you like to try?
anonymous
  • anonymous
yes give me a min to do it
anonymous
  • anonymous
okay
anonymous
  • anonymous
|dw:1440524508903:dw|
anonymous
  • anonymous
excuse the terrible penmanship lol
anonymous
  • anonymous
very good!!!
anonymous
  • anonymous
so now we have \[\Large{\frac{ \frac{ 2x }{ x+2 } }{ \frac{ 5x+20 }{ x+2 } }}\] right?
anonymous
  • anonymous
|dw:1440524931675:dw|
anonymous
  • anonymous
and then i do this?
anonymous
  • anonymous
yes, do you know why?
anonymous
  • anonymous
because its like dividing regular fractions... you flip and multiply... sooo the (x+2) crosses out right?
anonymous
  • 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
  • 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
  • anonymous
does that make sense?
anonymous
  • anonymous
yes so you factor 5x+20... to get 5(x+4)
anonymous
  • anonymous
\[\frac{ 2x }{ 5(x+4)}\] is the answer
anonymous
  • anonymous
or should i not factor it?
anonymous
  • anonymous
oh waiiittt hold on
anonymous
  • anonymous
wait nvm thats my final answer lol
anonymous
  • anonymous
to factor or not to factor is a matter of preference... you still have reduced to lowest terms.
anonymous
  • anonymous
but does the process make sense to you?
anonymous
  • 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
  • anonymous
yep
anonymous
  • anonymous
good job!
anonymous
  • 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
  • anonymous
sure
anonymous
  • anonymous
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anonymous
  • anonymous
and i think that can be reduced?
anonymous
  • anonymous
correct, it can be reduced
anonymous
  • anonymous
well this is a silly brain fart but i would reduce all the terms by 2?
anonymous
  • anonymous
\[\frac{ 4x-12 }{ 5x}\]
anonymous
  • anonymous
yep, that's the extent of it
anonymous
  • anonymous
one more question lol
anonymous
  • anonymous
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anonymous
  • anonymous
for the first question the denominator isnt a fraction...
anonymous
  • anonymous
you can make it one... \[5+a=\frac{ 5+a }{ 1 }\]
anonymous
  • anonymous
ohhhh wait i got thiis
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anonymous
  • anonymous
yes except it is \(25-a^2 \text{ not } a^2 - 25\)
anonymous
  • anonymous
gotchaaa.. but does the order i write it affect it? i thought it was a difference of squares either way
anonymous
  • anonymous
true, but \(25-a^2 \ne a^2-25\) unless \(a=\pm 5\)
anonymous
  • anonymous
ahh i gotchaa. seriously thank you so much! youre amazingg! i might message you when im in need of a lesson lol
anonymous
  • anonymous
right on! good luck, have a great day and keep on learning!

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