anonymous
  • anonymous
Help solve please medal! Simplify 3x+10/ x+1 + x+3/x+4 -6x+27/(x+1)(x+4)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@radar @sammixboo @MrNood @puppylife101
anonymous
  • anonymous
@Yungwisdom2000 @Compassionate
jtvatsim
  • jtvatsim
I'm going to rewrite this to make sure we are looking at the same thing. :)

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anonymous
  • anonymous
ok thanks :)
jtvatsim
  • jtvatsim
\[\frac{3x+10}{ x+1} + \frac{ x+3}{x+4} - \frac{6x+27}{(x+1)(x+4)}\] Is that the right thing?
anonymous
  • anonymous
correct
jtvatsim
  • jtvatsim
It's a bit scary looking. :) Any thoughts on what you might do first?
anonymous
  • anonymous
we need a common denominator right :)
jtvatsim
  • jtvatsim
That would be REALLY nice. Yes, let's try that.
jtvatsim
  • jtvatsim
What might be the common denominator we are trying to make?
anonymous
  • anonymous
(x+1) (x+4) right
jtvatsim
  • jtvatsim
Seems like a good choice. That appears to be the most complicated thing in our expression. So far, so good.
jtvatsim
  • jtvatsim
So, let's look at the first fraction. What needs to be done here?
anonymous
  • anonymous
3x+10 (x+4) / x+1 (x+4) + (x+3) (x+1) / (x+4)(x+4) right
jtvatsim
  • jtvatsim
Yes, on the second fraction you mean (x + 1) on the bottom, but I get it. :)
jtvatsim
  • jtvatsim
Great! So here's what we have so far.
jtvatsim
  • jtvatsim
\[\frac{(3x+10) (x + 4)}{(x+1) (x + 4)}+\frac{(x+3)(x+1)}{(x+1)(x+4)}−\frac{6x+27}{(x+1)(x+4)}\]
jtvatsim
  • jtvatsim
Wow, that's a lot of stuff. Now what?
anonymous
  • anonymous
cancel out
jtvatsim
  • jtvatsim
Well, if we "cancel out" that might put us right back where we started right? Unless you mean distribute...
anonymous
  • anonymous
i'm sorry i meant distribute lol
jtvatsim
  • jtvatsim
No worries. :) Let's try to do that. What do you get. (Might take a minute) :)
anonymous
  • anonymous
im kinda lost can you draw it out
anonymous
  • anonymous
from where we left off
jtvatsim
  • jtvatsim
Sure! I think we were here right? \[\frac{(3x+10)(x+4)}{(x+1)(x+4)}+\frac{(x+3)(x+1)}{(x+1)(x+4)}−\frac{6x+27}{(x+1)(x+4)}\]
anonymous
  • anonymous
yes ok, so we distribute
jtvatsim
  • jtvatsim
Yes, and what do you get after that?
anonymous
  • anonymous
i got 4/ (x+1) (x+4)
jtvatsim
  • jtvatsim
4 for which fraction?
jtvatsim
  • jtvatsim
As the final answer?
anonymous
  • anonymous
yes the final
anonymous
  • anonymous
i think i did something wrong
jtvatsim
  • jtvatsim
Oh, ok, let me double check that. :)
anonymous
  • anonymous
yes please check
jtvatsim
  • jtvatsim
So, for the first fraction: \[\frac{(3x+10)(x+4)}{(x+1)(x+4)}=\frac{3x^2 + 12x + 10 x + 40}{(x+1)(x+4)}\]
jtvatsim
  • jtvatsim
For the second fraction: \[\frac{(x+3)(x+1)}{(x+1)(x+4)} = \frac{x^2 + 3x + x + 3}{(x+1)(x+4)}\]
jtvatsim
  • jtvatsim
For the third fraction (watch out for the tricky - sign!): \[-\frac{6x+27}{(x+1)(x+4)}=\frac{-6x-27}{(x+1)(x+4)}\]
anonymous
  • anonymous
ok i got 1/ (x+4)(x+1) for final
jtvatsim
  • jtvatsim
Checking again... :)
jtvatsim
  • jtvatsim
I get: (3x^2 + x^2 + 22x + 4x - 6x + 40 + 3 - 27)/(x+1)(x+4) so (4x^2 + 20x + 16)/(x+1)(x+4) so (2x + 2)(2x + 8)/(x+1)(x+4) then 2(x+1)*2(x+4)/(x+1)(x+4) Oh wow! 4 is the final answer. I think. :)
anonymous
  • anonymous
ok i see, now you factor
anonymous
  • anonymous
hey do you mind checking this one for me
anonymous
  • anonymous
x/3 div 2/x+5
anonymous
  • anonymous
i got X(x+5) /6

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