Algebra 2 help please!! 1. What is the simplified form of (x-3)/(x^2+x-12)• (x+4)/(x^2+8x+16) ? 2.What is the simplified form of (x+1)/(x^2+x-6) divided by (x^2+5x+4)/(x+3)? 3.What is the simplified form of (y)/(y^2-x^2) – (x)/(y^2-x^2) ? 4. What is the simplified form of (2)/(x^2-x) - (1)/(x) ? Answers I got: 1. (1)/(x+4)^2 2.(1)/(x-2)(x+4) 3.(1)/(x-y) 4.(1-x)/(x(x-1)) Thanks!!
1. - Correct
2. - Correct
3. I got: y/(y+x)(y-x) - x/(y+x)(y-x) = (y-x)/((y+x)(y-x)) = 1/(y+x)
Thanks so muich for all of the help!! You rock!
4. For this one, I got: 2/x(x-1) - (x-1)/x(x-1) = (2-(x-1))/(x(x-1)) = (3-x)/(x(x-1))
You couldn't possibly tell me how to simplify this too could you?? (3)/(2x-5) + (21)(8x^2-14x-15) Thanks again!
I'll give it a shot :-)
Awesome!
That second one is going to be 21 over (2x+5)(4x-3). I just did it by trial and error. Wanna try solving from there and we'll compare?
Okay sounds good!
3(4x-3)/((2x+5)(4x-3)) + 21/((2x+5)(4x-3)) = (3(4x-3) +21)/((2x+5)(4x-3))
I think the final answer would be 6(x+1)/(2x+5)(4x+3)
I got (12(x+1))/((2x+5)(4x-3))
Okay, thanks so much!
Oh CRAP! I treated it like +14 in the polynomial
If you would like to go off and work on another problem, I'll resolve it for you.
Okay, thanks :)
= 3/(2x+5) + 21/((2x-5)(4x+3)) =(3(2x-5)(4x+3))/((2x+5)(2x-5)(4x+3)) + (21(2x+5))/((2x+5)(2x-5)(4x+3)) =(3(2x-5)(4x+3) + 21(2x+5))/((2x+5)(2x-5)(4x+3)) = ((6x-15)(4x+3) +42x +105)/((2x+5)(2x-5)(4x+3)) = (24x^2 -42 +42 +105)/((2x+5)(2x-5)(4x+3)) = (24x^2 +105)/((2x+5)(2x-5)(4x+3)) Pretty lengthy, I know :-(
Wow! Thanks so much!! You wouldn't happen to know how to solve fractions inside fractions would you?
from inside out?
Ya, I am trying that, do you want to give it a go as well of would you rather not?
sure I'll look at this last one.
Cool, thanks! what is the simplified form of (15xy^2)/(x^2+5x+6) over (15xy^2)/(x+2)(x+3)
So I'll ask you this. If you have \[a/b \div c/d\] What would you do to re-write it?
Change it to multiplication and flip the second fraction, would this one be the same though because of the fraction as a numerator and a fraction as the denominator?
Yup. So you have a/b * d/c
sec, I'll re-write what you have to give you a hint...
I wrote the wrong equation, it should be: what is the simplified form of (15xy^2)/(x^2+5x+6) over (5x^2y)/(2x^2+7x+3) Sorry!!
\[(15xy ^{2})/(x ^{2}+5x+16) \div (15xy ^{2})/((x+2)(x+3)\]
So if you flip the right fraction, you'll be able to cancel out 15xy^2.
Oh, it's 6 on the bottom, not 16. Sorry about that.
I wrote the equation wrong the first time, read the post right before you finished writing it all out. Sorry
oh okay one sec.
Do you know how to factor this? x^2+5x+6
Yes, I got (x+2)(x+3) but I do not know how to factor 2x^2+7x+3
Hmm... I don't think it's factor-able.
\[(15xy^2)/((x+2)(x+3)) \div (5x^2y)/(2x^2+7x+3)\]
It's not factorable, but that doesn't matter. What do you do to re-write this?
Multiply and flip the last fraction, that is where I got stuck
K, that's fine. I just wanted to make sure you were at that step.
Yep :)
For the final answer, I ended up with: (3y *(2x^2+7x+3))/((x+2)(x+3)*x)
The 15xy^2/5x^2y becamse 3y/x
Alrighty, thanks for the help
Some of these get pretty messy!
Yeah, they do!
If you can really learn all of this basic stuff, you are going to THANK yourself when you get into calculus.
You got that right!!!
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