Ask your own question, for FREE!
Mathematics 77 Online
OpenStudy (anonymous):

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!!

OpenStudy (anonymous):

1. - Correct

OpenStudy (anonymous):

2. - Correct

OpenStudy (anonymous):

3. I got: y/(y+x)(y-x) - x/(y+x)(y-x) = (y-x)/((y+x)(y-x)) = 1/(y+x)

OpenStudy (anonymous):

Thanks so muich for all of the help!! You rock!

OpenStudy (anonymous):

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))

OpenStudy (anonymous):

You couldn't possibly tell me how to simplify this too could you?? (3)/(2x-5) + (21)(8x^2-14x-15) Thanks again!

OpenStudy (anonymous):

I'll give it a shot :-)

OpenStudy (anonymous):

Awesome!

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

Okay sounds good!

OpenStudy (anonymous):

3(4x-3)/((2x+5)(4x-3)) + 21/((2x+5)(4x-3)) = (3(4x-3) +21)/((2x+5)(4x-3))

OpenStudy (anonymous):

I think the final answer would be 6(x+1)/(2x+5)(4x+3)

OpenStudy (anonymous):

I got (12(x+1))/((2x+5)(4x-3))

OpenStudy (anonymous):

Okay, thanks so much!

OpenStudy (anonymous):

Oh CRAP! I treated it like +14 in the polynomial

OpenStudy (anonymous):

If you would like to go off and work on another problem, I'll resolve it for you.

OpenStudy (anonymous):

Okay, thanks :)

OpenStudy (anonymous):

= 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 :-(

OpenStudy (anonymous):

Wow! Thanks so much!! You wouldn't happen to know how to solve fractions inside fractions would you?

OpenStudy (anonymous):

from inside out?

OpenStudy (anonymous):

Ya, I am trying that, do you want to give it a go as well of would you rather not?

OpenStudy (anonymous):

sure I'll look at this last one.

OpenStudy (anonymous):

Cool, thanks! what is the simplified form of (15xy^2)/(x^2+5x+6) over (15xy^2)/(x+2)(x+3)

OpenStudy (anonymous):

So I'll ask you this. If you have \[a/b \div c/d\] What would you do to re-write it?

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

Yup. So you have a/b * d/c

OpenStudy (anonymous):

sec, I'll re-write what you have to give you a hint...

OpenStudy (anonymous):

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!!

OpenStudy (anonymous):

\[(15xy ^{2})/(x ^{2}+5x+16) \div (15xy ^{2})/((x+2)(x+3)\]

OpenStudy (anonymous):

So if you flip the right fraction, you'll be able to cancel out 15xy^2.

OpenStudy (anonymous):

Oh, it's 6 on the bottom, not 16. Sorry about that.

OpenStudy (anonymous):

I wrote the equation wrong the first time, read the post right before you finished writing it all out. Sorry

OpenStudy (anonymous):

oh okay one sec.

OpenStudy (anonymous):

Do you know how to factor this? x^2+5x+6

OpenStudy (anonymous):

Yes, I got (x+2)(x+3) but I do not know how to factor 2x^2+7x+3

OpenStudy (anonymous):

Hmm... I don't think it's factor-able.

OpenStudy (anonymous):

\[(15xy^2)/((x+2)(x+3)) \div (5x^2y)/(2x^2+7x+3)\]

OpenStudy (anonymous):

It's not factorable, but that doesn't matter. What do you do to re-write this?

OpenStudy (anonymous):

Multiply and flip the last fraction, that is where I got stuck

OpenStudy (anonymous):

K, that's fine. I just wanted to make sure you were at that step.

OpenStudy (anonymous):

Yep :)

OpenStudy (anonymous):

For the final answer, I ended up with: (3y *(2x^2+7x+3))/((x+2)(x+3)*x)

OpenStudy (anonymous):

The 15xy^2/5x^2y becamse 3y/x

OpenStudy (anonymous):

Alrighty, thanks for the help

OpenStudy (anonymous):

Some of these get pretty messy!

OpenStudy (anonymous):

Yeah, they do!

OpenStudy (anonymous):

If you can really learn all of this basic stuff, you are going to THANK yourself when you get into calculus.

OpenStudy (anonymous):

You got that right!!!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!