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cutiecomittee123
 one year ago
How many solutions are there to this system of equations?
2x+y^2=33
x^2+y^2+2x=19
1,2,3, or 4???
cutiecomittee123
 one year ago
How many solutions are there to this system of equations? 2x+y^2=33 x^2+y^2+2x=19 1,2,3, or 4???

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0mathway.com quickmath.com i promise they help so much

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1are you allowed to substitued 1 into the other?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1you're suppose to try. i'm suppose to help.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1why not rearrange for x in the first equation, and sub it into the next equation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Notice 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? :)

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1this is another question.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1get that junk into another question.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It would be 0. Are you sure 0 isnt one of the answer choices?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Unless mobile is screwing with me that badly and the question i see is different than where Im typing :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Although you dont have to be so rude about it either way _

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1ah. sorry, i'm working on 0 hrs of sleep, and university work and job applications. forgive me.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright. 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?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1looking at too many problems at the same time.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, definitely 0 solutions. Cant copy the wolfram link on mobile. But what i did was perfectly fine and shows there is no solution

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1I worked through the entire problem and got something different.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, do you see what i did to come across my solution?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, went to wolfram. It made sense. I'm double checking my answer now.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=2x%2By%5E2%3D33+and+x%5E2%2By%5E2%2B2x%3D19 is what you went thru right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1Alright. I'm going to redo 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.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1I think I already found my problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, 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.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1I dunno. I worked thru it again, and I got 7.81=y.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you type out tour steps then, tim?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1Rearranged x= (y^2+33)/2

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1then i subsituted into the other equation

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1((y^2+33)/2)^2+y^2+2((y^2+33)/2)=19

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1(0.5y^2+16.5)^2+y^2y^2+33=19

NotTim
 one year ago
Best ResponseYou've already chosen the best response.10.5y^2+y^2y^2+16.5+3319=0

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1Which 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.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1@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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, i kost wifi. Ill explain where in your steps you can get no solutiom nottim

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0At 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wow, it deleted my wxplanatiom

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, it got deleted again. This time ill copy it in case that happens again, haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0At 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

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1You forgot to subtract 16.5^2 to both sides.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1But the order is 3 I think.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1wait wrong place what is that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im 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.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1(0.5y^2+16.5)^2 = 14.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1Unless you typed that wrong.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thats after i cancelled y^2's and subtracted 33.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1that would result in 0.25y^4+272.25= 14

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1then 0.25y^4+272.25=14

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, it wouldnt. (0.5y^2 +16.5)^2 = .25y^4 16.5y^2 +272.25

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1why does 16.5^2 become 16.5y^2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its a foiling problem if you were to expand it. That and squaring 0.5y^2 includes squaring the negative, it would disappear

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Bleh, sick of it deleting my messages. Stupid mobile, lol

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1Alright, you reasoning sounds perfect. Of course, I'm going to check once more, but everything you stated is sound.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But 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

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1Darnit i'm confused again.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1(0.5y^2+16.5)(0.5y^2+16.5)+33= 19 was a step?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1Alright, figured it out. Finally. Thanks mate.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Haha, okay. As long as everything makes sense. Do you come to no solution now?

NotTim
 one year ago
Best ResponseYou've already chosen the best response.1Same problem I had in high school. Didn't check over my work.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lol, alright. Well, cutie closed the question, so not sure what conclusion she came to or if she made another question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, sick of my wifi. Ill log back on once im in a position to set up my laptop. Laterz :)
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