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Algebra II



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Mon Mar 16, 2009 7:45 pm
Rosendorn says...



Given- x+ y= Irrational 2 (don't know how to do symbols like that), x2+ y2= 6

The problems I need to solve with a formula like that are:

xy= ______

x3+ y3= ______

y/x+ x/y= ______

y/x2+ x/y2= ______

I'd like a step-by-step thing, if possible. Mostly on the last three problems.

Thank you in advance!
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Tue Mar 17, 2009 12:17 am
fluteluvr77 says...



'Kayy Rosey, I'm assuming Irrational 2 is √2 and that x2 is x squared, which I write as x^2...If not, please correct me 'kay? 'Cause that's what my teacher uses...

So, we have :
x + y = √2
x^2 +y^2 = 6

I'll name the first equation 1 (x+y = √2) and the second (x^2+y^2 = 6) equation two (sorry about the formalities, I just tend to find this easier when dealing with a lot of equations and stuff...)

Part A
Now, lets square the equation 1 to get rid of the radical.
(x+y)^2 = (√2)^2
x^2+y^2+2xy = 2 --> Equation 3

Now, we have that x^2 + y^2 = 6 from equation 2, so lets plug it in equation 3
(x^2+y^2) + 2xy = 2

Substitute 6 for (x^2 + y^2)
6+2xy = 2

Then, subtract 6 from both sides
2xy = -4

Divide by 2 to get :
xy = -2

Part B
Next we want x^3 + y^3

So, we have:
x+y = √2 --> Equation 1
x^2+y^2 = 6 --> Equation 2
xy = -2 -->Equation 3

So, lets cube the first equation to get
x^3+ 3x^2y + 3xy^2 + y^3 = 2√2

Now, we want to factor this into terms that we know, and what we wish to find.
So, lets leave the x^3 + y^3 our desired answer to the side
(x^3 + y^3) + (3x^2y+3xy^2) = 2√2

Now, since we know xy lets factor 3xy out from the second set of parantheses
(x^3+y^3) + 3xy(x+y) = 2√2

Look at that! We know both x+y (Eq. 1) and 3xy (Eq. 3 multiplied by 3)!
So, lets plug those in!

(x^3 + y^3) + [3(-2)](6) = 2√2
x^3 + y^3 -36 = 2√2
x^3 + y^3 = 36 + 2√2
Tada! Answer to part b.

Sorry, I gotta go now, but I'll be back soon to help with the last two problems! Hope I helped, and my explanations are clear!
fluteluvr77<3
Last edited by fluteluvr77 on Tue Mar 17, 2009 12:48 am, edited 1 time in total.
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Tue Mar 17, 2009 12:41 am
Rosendorn says...



Okay, you have to understand that my math is self-taught so the likelihood I know proper terminology is slim.

You're right about what the symbols are.

To solve xy, I have these intermediate steps in my textbook:

xy= (x+y)^2 - (x^2+ y^2)/2 = 2- 6/2= -2

(Word translation- x plus y squared minus x squared plus y squared over two)

And, which equation are you solving for? "Equation three" makes me think you're solving y/x+ x/y= ______

Can you please put some spaces between lines so I can read them easier? A block of text like that is very hard for me to see properly.
A writer is a world trapped in a person— Victor Hugo

Ink is blood. Paper is bandages. The wounded press books to their heart to know they're not alone.
  





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Tue Mar 17, 2009 12:44 am
fluteluvr77 says...



*Edit* Sorry Rosey I didn't see the post :oops: *
Back Rosey! With the last two answers. These are a lot easier than they look don't worry! ;)

So, to recap, we have :
x+y = √2 --> Equation 1
x^2+y^2 = 6 --> Equation 2
xy = -2 -->Equation 3
and our newest equation
x^3 + y^3 = 36 + 2√2 --> Equation 4

Part C
Now, we would like to find :
y/x +x/y

Let's start by finding a common denominator to make it one fraction. The LCM of x and y is xy, so we'll multiply the first term by y and the second by x.

So, we get
(y^2)/(xy)+(x^2)/(xy)
which, simplified, is :
(x^2+y^2)/(xy)

Now, we know all these values, so it's basically Eq. 2/Eq.3
So, plug it in to get :
6/-2 = -3

Part D
So, we would like to find
y/x^2 + x/y^2

So, again, let's take the common denominator which is x^2y^2.

We multiply the first equation by y^2 and the second by x^2, and get :
y^3/x^2y^2 +x^3/y^2x^2

Simplify it to get
(x^3+y^3)/(x^2y^2)

The numerator should look familiar! It's equation 4! The denominator isn't too hard to figure out since it's Equation 3 squared.

So, let's square equation 3!
xy = -2
(xy)^2 = (-2)^2

x^2y^2 = 4 --> Equation 5

Now, it's simply Eq. 4/Eq. 5 which is :
(36 + 2√2)/4

This simplified gives you
9+(1/2√2)

All done! Hope this helped you! Hehe, if I made a mistake, PM me or something 'kay? I'm mathematically challenged xD...
fluteluvr77<3
Last edited by fluteluvr77 on Tue Mar 17, 2009 12:53 am, edited 2 times in total.
Love is the answer to life yet the slowest form of suicide.
Love is a paradox.
And that's why we love it.

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Tue Mar 17, 2009 12:46 am
Rosendorn says...



I have a request in the post right above yours: Please put a space between lines so I can read them.
A writer is a world trapped in a person— Victor Hugo

Ink is blood. Paper is bandages. The wounded press books to their heart to know they're not alone.
  





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Tue Mar 17, 2009 12:52 am
fluteluvr77 says...



Is that better? Sorry about that...I thought YWS formats!
Love is the answer to life yet the slowest form of suicide.
Love is a paradox.
And that's why we love it.

Got YWS?
  





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Gender: Other
Points: 89625
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Tue Mar 17, 2009 1:00 am
Rosendorn says...



That is so weird. My textbook doesn't give that formula for solving those problems at all.

Ah well. Thanks for trying! I'll have to dig up a teacher for this program and ask them.

Thanks again!
A writer is a world trapped in a person— Victor Hugo

Ink is blood. Paper is bandages. The wounded press books to their heart to know they're not alone.
  








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