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mathway.com quickmath.com i promise they help so much
are you allowed to substitued 1 into the other?
you can try
you're suppose to try. i'm suppose to help.
why not re-arrange for x in the first equation, and sub it into the next equation.
Notice both equations have y^2 +2x in them. The first equation tells us this expression is equal to 33, so you can replace y^2 +2x for 33 in the second equation. Doing this gives x^2 +33 = 19 or x^2 = -14 How many solutions does that have? :)
this is another question.
get that junk into another question.
It would be 0. Are you sure 0 isnt one of the answer choices?
Unless mobile is screwing with me that badly and the question i see is different than where Im typing :/
Although you dont have to be so rude about it either way -_-
ah. sorry, i'm working on 0 hrs of sleep, and university work and job applications. forgive me.
Alright. So assuming mobile may be messing with me, i see the question as asking for how many solutions there are to the system 2x + y^2 = 33 X^2 + y^2 +2x = 19 Am i seeing the wrong question because of mobile?
nah, is good.
looking at too many problems at the same time.
Okay, then i stand by what i said. Her answer choices dont include 0 solutions. I can check wolfram to back me up, but might be slow on mobile, lol
Why is that?
Yeah, definitely 0 solutions. Cant copy the wolfram link on mobile. But what i did was perfectly fine and shows there is no solution
I worked through the entire problem and got something different.
Well, do you see what i did to come across my solution?
Yeah, went to wolfram. It made sense. I'm double checking my answer now.
http://www.wolframalpha.com/input/?i=2x%2By%5E2%3D33+and+x%5E2%2By%5E2%2B2x%3D19 is what you went thru right.
Yes, thats what i saw when i checked. Well, i can either try to better explain what i did or try and see where you had a mistake
Alright. I'm going to re-do my thing. If i do the same thing, I'll post it. If my results are in agreement, I'll tell ya. give me 3 minutes.
I think I already found my problem
Okay, thats good. Either way, were trying to make a substitution that lets us reduce one of the equations to a single variable. Thankfully, we have a common y^2 + 2x in each equation. The first equation explicitly tells us that y^2 + 2x = 33. This will ways be true. Because the 2nd equation contains the same y^2 +2x, we can replace that chunk with 33. x^2 +2x + y^2 = 19 becomes x^2 + 33 = 19. And then from there you can see the no solution result.
I dunno. I worked thru it again, and I got 7.81=y.
We could also show the same result by an elimination method, which is essentially going to be the same thing. Whichever way makes the result more visible to ya.
Can you type out tour steps then, tim?
Rearranged x= (-y^2+33)/2
Also, im NOT tim
then i subsituted into the other equation
Which doesn't make sense in the presence of wolfram alpha. I get that wolfram alpha sometimes get wrong data in context, but my written out answer doesn't collide well with that data.
@cutiecomittee123 In my conclusion, I believe it is 1. But don't trust me. both of us got different answers. But what you can take out of this conversation, is to rearrange 2x+y^2=33 into x=____ and then sub it into the other equation for x.
Sorry, i kost wifi. Ill explain where in your steps you can get no solutiom nottim
At the step youwhere you have (-0.5y^2 +16.5)^2 +y^2 -y^2 +33 = 19. So let me cancel the y^2's and move the 33 to the other side. That would give (-0.5y^2 +16.5)^2 = -14 Notice we have something squares equal to a negative value. This isnt possible and from here we can also conclude no solution
Wow, it deleted my wxplanatiom
Let me retype
So, the step where you have (-0.5y^2 +16.5)^2 +y^2 -y^2 +33 = 19. Let me cancel the y^2's and move the 33 to the other side. This gives (-0.5y^2 +16.5)^2 = -14. Notice how we have a squared term equal to a negative value. From here we can see the no solution result @NotTim
Well, it got deleted again. This time ill copy it in case that happens again, haha
At the step where you have (-0.y^2 + 16.5)^2 +y^2 - y^2 + 33 = 19 Ill cancel the y^2's and move the 33 to the other side. This gives (-0.5y^2+16.5)^2 = -14. Thus we have something squared equal to a negative value. This shows that we would have no solution since a value squared cannot equal something negative
You forgot to subtract 16.5^2 to both sides.
But the order is 3 I think.
wait wrong place what is that
Im just looking at your steps and i dont see where that would be. There wouldnt be a 16.5^2 to subtract. You have the entire quantity of (-0.5y^2 +16.5)^2, i cannot subtract 16.5^2 from here.
(-0.5y^2+16.5)^2 = -14.
Unless you typed that wrong.
Thats after i cancelled y^2's and subtracted 33.
that would result in -0.25y^4+272.25= -14
No, it wouldnt. (-0.5y^2 +16.5)^2 = .25y^4 -16.5y^2 +272.25
why does 16.5^2 become -16.5y^2?
Its a foiling problem if you were to expand it. That and squaring -0.5y^2 includes squaring the negative, it would disappear
Bleh, sick of it deleting my messages. Stupid mobile, lol
Alright, you reasoning sounds perfect. Of course, I'm going to check once more, but everything you stated is sound.
But okay, lets write out each individual step of the foil -0.5y^2 * -0.5y^2 = 0.25y^4 -0.5y^2 * 16.5 = -8.25y^2 16.5 * -0.5y^2 = -8.25y^2 16.5 * 16.5 = 272.25 put it together and you have 0.25y^4 -16.5y^2 + 272.25
Darnit i'm confused again.
(-0.5y^2+16.5)(-0.5y^2+16.5)+33= 19 was a step?
Alright, figured it out. Finally. Thanks mate.
Haha, okay. As long as everything makes sense. Do you come to no solution now?
Same problem I had in high school. Didn't check over my work.
Lol, alright. Well, cutie closed the question, so not sure what conclusion she came to or if she made another question
Alright, sick of my wifi. Ill log back on once im in a position to set up my laptop. Laterz :)