Math question! please help..

[Nynaeve]

x^2 + y = 41

y^2 + x = 31

 

 

what's x and y? 

 

show me the way, please. if you do know.

 

 

 

Nyn

[SBoydW]

ack.... letters and amth don't mix.... :-X

[MsDanya]

now thats exactly why I stopped taking math many years ago  :P

[Cadsuane]

Well, you could graph the Y equations on the same graph and find the

point(s) where the two lines intersect, and those are your solution(s). Because of the exponents, they won't be straight lines, so there may be more than one point of intersection, and thus more than one correct answer.

 

I think the graphs would look like

a) an upside down U with the base at the point (0,41), and

b) a sideways U, open end pointing left, with the base at the point (31,0).

 

That means the solution set will probably be four points in the form of (x,y).

 

 

To do it mathematically is ugly, and I can't remember how to solve the last bit of what I get. If someone else here can solve

 

1650 + x - 82(x^2) + x^4 = 0

 

for the four (I think) x values that will make it true, then I'll post how I got that equation mathematically. Otherwise, it might be useless.  :-\

[loki redfern]

er i found that y=5, x=6 works ..but dont ask how ..er it wasnt very ...mathsy  :-[ ..well trial and error is  actuall but yeh so other than just trying  out the first 5 or 6 numbers it could be and seeing which work in both  ( i was being to lazy to check for -ve solutions) that one had me, ususally you try to make them equal to one another, or sub one in to the other..but those are nasty equns ..or you can try solving grapgically, er that might work..

 

meh  ok after seeing cads post i  can show the graph (trial and error triumps again :D)

 

this shows the positive co-ooridinates solutions,

 

http://i52.photobucket.com/albums/g2/iriejay/g-ve.png[/img]

this one the nevative which is like x=-5.91(-5.9097) , y=6.08(6.0753 to 3sf.

 

 

so;

x=5, -5.91

y=6,  6.08

 

oh aye , yeh ou mahe y the subject in each to get the the equations you use

 

erm i think there arent 4 equations becuae one will be the line for one has a square root in its formula so the are no real -ve part to that one (and hence none is below y=0)(the horizontally one..er yeh i think ill leave explining and probable correcting to cads she sounds er more able ;D 

 

 

[Cadsuane]

No, I think there are 4 points. I do vaguely remember something tying a square root to only having a positive answer, but I don't think that it is applicable here, because you are deriving the square root from a squared number, which means that it is actually +/- sqrt, not just +sqrt, as it would be if you were given it that way without an (-) indicator. When you take the square root of something, the answer is both the positive and negative square roots, not just the positive.

 

Additionally, I think the fact that the equation I posted above (and again below, here) has an x^4 is an indication that there are four X values that will make it true.

 

1650 + x - 82(x^2) + x^4 = 0

 

 

The idea to solve it would be to break that equation down into something that looks like

 

(__ +/- __)*(__ +/- __)*(__ +/- __)*(__ +/- __) = 0

 

(a simplified example would be (x+1)(x-2)(x+3)(x-4) = 0)

 

where each (__ +/- __) has an x as one of the blanks. And then, because anything times 0 is 0, then whatever value for X makes the value of that single (__ +/- __) equal to zero, would be one of the four values of x that make the equation true, and thus would be the x value for one of the four points. The solution for the example above would be X values {-1,2,-3,4}.

 

Then, you'd plug the four X values into one of the two equations (I'd choose [y = 41 - x^2], as it's not got the complicating sqrt in it), and get the four solutions for Y.

 

I think. ;D

[Nynaeve]

Thing is...I knew the answers.

 

But I need to show the exact way of getting it.... bah!

 

 

 

 

Nyn

[Deleios]

Just a little detail: There are at most 4 points, it doesn't mean that there has to be 4 solutions.

 

The probleme is same as here, as we can state that 6 is an "obvious" solution, the factorization gives this:

 

(x-6)(x^3+6x^2-46x-275)=0

 

So now, let's solve x^3+6x^2-46x-275=0 ^^

 

Q(x)=x^3+6x^2-46x-275

Q'(x)=3x²+12x-46

 

Delta= 12²- 4*3*(-46)=696>0

 

Q'(x1)=0 Q'(x2)=0

 

x1=(-12-sqrt(696))/6 <0  and x2=(-12+sqrt(696))/6 >0

 

Obviously, Q'>0 when x<x1 or x>x1, which means Q grows between ]-∞; x1] and [x2;+∞[

 

(If I'm not mistaken, Q(x1)>Q(x2))

 

So if Q(x1)<0, that means you only have one other solution.

If Q(x1)>0 and Q(x2)>0 you only have one too.

If Q(x1)>0 and Q(x2)<0 then there are 3 solutions.

 

I'll leave you the rest, ask if you did not understand my explanation ^^

 

 

 

 

 

 

[Cadsuane]

Why would it not have 4 points, if the graph looks like this?

 

 

 

 

Also, how did you get/choose (x-6) to take out? I know it's correct, I just can't remember how to do the factorization. If you don't want to spend time to remind me for no reason, don't. I won't mind. ^_^

[Deleios]

Because 6-6=0, and 6 is a solution ;)

 

I have edited my post with further explanation

 

Besides, x²+x+1=0 doesn't have any solution (and from what you said, you could think there are two). Well, no real solution at last, so it is no sure thing that there are 4 real solutions. But there are at most 4.

[loki redfern]

isnt the one with the sideways curve only exist at y>/=0, the equation for it is er y=(31-x)^1/2  (which is what i put in to my graph drawing thing, and ^1/2 = square rooting and a square root only gives +ve values as the -ve ones come from ...er imaginary numbers or something...i could be totally wrong though and so i if i apologie like

 

ow its really getting to late for my brain to be working

[loki redfern]

*looks round to check normal people arnt around* this is actually quite fun, we should do it again some time ;D

[Deleios]

Yep lol :)

 

If I'm not mistaken (I hope I'm not)

 

x1=-6,4

x2=2.4

 

 

[Cadsuane]

Hah! I remember doing that stuff in high school (maybe college, too... everything after the differential and integral calculus classes gets blurry ^_^), but it's beyond me these days. Thanks, though.  :D

 

Edit: Gotcha, Loki. My memory is failing, surprise surprise. ^_^

[Deleios]

I did that a couple of years ago, too, and I'm quite upset that I don't remember that well, I used to be good at this.

 

Well, while I'm at it...

 

Q(x1)=3

Q(x2)=-337

 

Which means there are 3 other solutions, solution2<x1, x1<solution3<x2, x2<solution3

 

How to find them?

 

I don't have a clue!  ;D

[Lord_Naryb]

Hey, just found this! ;D

 

Okay here is how to do it.

 

x^2 + y = 41                          y^2 + x = 31

 

solve for y in the second equation y^2 = 31 - x

                                  y = (31-x)^(1/2)

 

plug that into the first equation

 

x^2 + [(31-x)^(1/2)] = 41

(31-x)^(1/2) = 41 - x^2

 

square both sides

 

31 - x = (41 - (x^2))^2

31 - x = 1681 - 41x^2 - 41x^2 + x^4

 

rearranging

 

x^4 - 82x^2 + x + 1650 = 0

 

Solve for x (I used my calculator to solve it (TI-85))

 

x1 = -6.8668

x2 = 6.7766

x3 = 6

x4 = -5.9097

 

Plug those back into the original equations

 

y1 = -6.1529

y2 = -4.9223

y3 = 5

y4 = 6.0754

 

Hope that helps   

[Zardi]

Nyn, are you allowed to use technology to solve this (like everyone has done so far), or do you have to explain how to do it by hand?

 

Here's a way to do it by hand:

 

x^2 + y = 41                y^2 + x = 31

 

Write the second one as x = 31 - y^2

Then plug that into the first one to get: (31-y^2)^2 + y = 41

...which simplifies to: y^4 - 62y^2 + 31^2 + y - 41 = y^4 - 62y^2 + y + 920 = 0

 

Then, you look at the factors of 920: 2, 5, and 23, and you try dividing the polynomial by (y - factor) or (y + factor). In this case, you can divide by (y - 5), so then you get:

 

(y-5)(y^3 + 5y^2 - 37y - 184) = y^4 - 62y^2 + y + 920 = 0

 

This shows you that the two graphs intersect at y = 5. Plug that into one of your original equations to get x = 6, so the first solution you find is (6,5).

 

Now, you can use any method of solving cubic equations to solve y^3 + 5y^2 - 37y - 184 = 0. You may have learned some in your class, but if not, try here: http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method

 

You might still need a basic calculator to approximate the fractions (they will get very messy otherwise), but at least you won't need a graphing calculator to do all the work for you.