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

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

Mathematics
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I would attempt to eliminate the x^2 term and solve for y first then go back and solve for x
notice if you multiply the second equation by -2 you will be able to then add the equations together to eliminate the x^2 term
I did that but I got y=sqrt5

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then I plugged it back into the first equation to solve for x and it came out as x=15.4
I'm not seeing how you obtained that
can you should your work so I can see how you first obtained y
If 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)
2x^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
how did you get sqrt(5)?
After adding the equations I get -y^2=-5
2x^2+y^2=33 x^2+y^2+2y=19 first step multiply second equation by -2 2x^2+y^2=33 -2x^2-2y^2-4y=-38 now add the equations y^2-2y^2-4y=33-38 simplify -y^2-4y=-5 multiply both sides by -1 to make it prettier y^2+4y=5 now subtract 5 on both sides y^2+4y-5=0
I got square root 5 because I multiplied both sides by -1 and then took the square root to get y=sqrt5
So y^2 +4y=5
you can factor y^2+4y-5
Help me with that?
what two numbers multiply to be -5 and add up to be 4?
think -5 can be written as the following: -5(1) or 5(-1) -5+1=? 5+(-1)=?
-4 or 4
so which pair of numbers satisfy my question?
"what two numbers multiply to be -5 and add up to be 4?"
oh 5 and -1
y^2+4y-5=0 (y+5)(y-1)=0
(x+5) (y-1) = 0
so now you just set both factor equal to 0 y+5=0 or y-1=0
y=-5 and or y=1
sounds 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
x=2
you 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
actually you might end up with 4 solutions since x has a square on it
yep there is one more solution to x^2=4 I assume that is where you got x=2 from
x also =-2
so we have (2,-5) and (-2,-5) now we need to solve for x when y is 1
you can still use 2x^2+y^2=33
uh y=sqrt31
did you replace x with 1?
yeah
remember 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
in other words we never said anything about x being 1
\[2x^2+y^2=33 \\ 2x^2+1^2=33 \\ 2x^2+1=33\] solve for x
2(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
why are you replacing x with 1? we never said anything about x being 1
Oh lol, I get x=4 after setting y=1
or also?
I assume you got that from x^2=16
but there is one more solution to that
x=-4
\[(2,-5); (-2,-5) ; (-4,1);(4,1)\]
sweet
okay now the question is which one of these is in quadrant 4 of the graph?
the 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)
|dw:1433531008630:dw|
(2,-5)
|dw:1433531053701:dw|
yep sounds great
Yes:)
lol this again
|dw:1433531133632:dw|
Can you help me with something else?:)
I can try
Okay its pretty simple what is the maximum number of solutions to this system x^2+4y=64 and x+y=5
well one is a parabola and one is a line
what is the max number of times a line can intersect a parabola?
2 times
yep 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.
Awesome thanks
I have more but I think I am going to ask them in other questions so things dont get confusing
k
Wait so i was right in the other question? Dude convinced me otherwise. goddarnit.
wait no, here it was 2. i only got 1.
what other question?>

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