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cutiecomittee123
 one year ago
How do I solve this system of equations
2x^2+y^2=33
x^2+y^2+2y=19
cutiecomittee123
 one year ago
How do I solve this system of equations 2x^2+y^2=33 x^2+y^2+2y=19

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freckles
 one year ago
Best ResponseYou've already chosen the best response.4I would attempt to eliminate the x^2 term and solve for y first then go back and solve for x

freckles
 one year ago
Best ResponseYou've already chosen the best response.4notice if you multiply the second equation by 2 you will be able to then add the equations together to eliminate the x^2 term

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0I did that but I got y=sqrt5

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0then I plugged it back into the first equation to solve for x and it came out as x=15.4

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I'm not seeing how you obtained that

freckles
 one year ago
Best ResponseYou've already chosen the best response.4can you should your work so I can see how you first obtained y

freckles
 one year ago
Best ResponseYou've already chosen the best response.4If you wouldn't mind what do you get right after you add the two equations (you know after the second one has been multiplied by 2)

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.02x^2+sqrt5=33 sqrt5 sqrt5 2x^2=30.76 divide by 2 on both sides and take the square root! (I forgot about that) So it would be x=sqrt 15.38

freckles
 one year ago
Best ResponseYou've already chosen the best response.4how did you get sqrt(5)?

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0After adding the equations I get y^2=5

freckles
 one year ago
Best ResponseYou've already chosen the best response.42x^2+y^2=33 x^2+y^2+2y=19 first step multiply second equation by 2 2x^2+y^2=33 2x^22y^24y=38 now add the equations y^22y^24y=3338 simplify y^24y=5 multiply both sides by 1 to make it prettier y^2+4y=5 now subtract 5 on both sides y^2+4y5=0

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0I got square root 5 because I multiplied both sides by 1 and then took the square root to get y=sqrt5

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you can factor y^2+4y5

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0Help me with that?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4what two numbers multiply to be 5 and add up to be 4?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4think 5 can be written as the following: 5(1) or 5(1) 5+1=? 5+(1)=?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so which pair of numbers satisfy my question?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4"what two numbers multiply to be 5 and add up to be 4?"

freckles
 one year ago
Best ResponseYou've already chosen the best response.4y^2+4y5=0 (y+5)(y1)=0

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0(x+5) (y1) = 0

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so now you just set both factor equal to 0 y+5=0 or y1=0

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0y=5 and or y=1

freckles
 one year ago
Best ResponseYou've already chosen the best response.4sounds great now you can go back and find x so use either equation Let's just use the first one 2x^2+y^2=33 replace y with 5 then solve for x then after you have completed that step replace y with 1 then solve for x

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you will have two different solutions two different ordered pairs (x,5) and (x,1) where you have to solve for both of those x's

freckles
 one year ago
Best ResponseYou've already chosen the best response.4actually you might end up with 4 solutions since x has a square on it

freckles
 one year ago
Best ResponseYou've already chosen the best response.4yep there is one more solution to x^2=4 I assume that is where you got x=2 from

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so we have (2,5) and (2,5) now we need to solve for x when y is 1

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you can still use 2x^2+y^2=33

freckles
 one year ago
Best ResponseYou've already chosen the best response.4did you replace x with 1?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4remember we had y=5 or y=1 this is why we are trying to find when x when y=5 and also why we are trying to find x when y=1

freckles
 one year ago
Best ResponseYou've already chosen the best response.4in other words we never said anything about x being 1

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[2x^2+y^2=33 \\ 2x^2+1^2=33 \\ 2x^2+1=33\] solve for x

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.02(1)^2+Y^2=33 2(1)^2=2 RIGHT? then 2+y^2=33 minus 2 from both sides then take the square root so y=sqrt31

freckles
 one year ago
Best ResponseYou've already chosen the best response.4why are you replacing x with 1? we never said anything about x being 1

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0Oh lol, I get x=4 after setting y=1

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I assume you got that from x^2=16

freckles
 one year ago
Best ResponseYou've already chosen the best response.4but there is one more solution to that

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[(2,5); (2,5) ; (4,1);(4,1)\]

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0okay now the question is which one of these is in quadrant 4 of the graph?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4the first two pairs come from replacing the y with 5 and solving for x (got x=2 or 2) and the last two pairs come from replacing the y with 1 and solving for x (got x=4 or 4)

freckles
 one year ago
Best ResponseYou've already chosen the best response.4dw:1433531008630:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.4dw:1433531053701:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.4dw:1433531133632:dw

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0Can you help me with something else?:)

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0Okay its pretty simple what is the maximum number of solutions to this system x^2+4y=64 and x+y=5

freckles
 one year ago
Best ResponseYou've already chosen the best response.4well one is a parabola and one is a line

freckles
 one year ago
Best ResponseYou've already chosen the best response.4what is the max number of times a line can intersect a parabola?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4yep so without solving the system we could say there is a max number of solutions to the system above of 2 and there is minimum number of 0 solutions so you can you have 2 solutions,1 solution, or 0 solutions to a system containing a parabola and a line.

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0Awesome thanks

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0I have more but I think I am going to ask them in other questions so things dont get confusing

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0Wait so i was right in the other question? Dude convinced me otherwise. goddarnit.

NotTim
 one year ago
Best ResponseYou've already chosen the best response.0wait no, here it was 2. i only got 1.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4what other question?>
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