cutiecomittee123
  • cutiecomittee123
How do I solve this system of equations 2x^2+y^2=33 x^2+y^2+2y=19
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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freckles
  • freckles
I would attempt to eliminate the x^2 term and solve for y first then go back and solve for x
freckles
  • freckles
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
cutiecomittee123
  • cutiecomittee123
I did that but I got y=sqrt5

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cutiecomittee123
  • cutiecomittee123
then I plugged it back into the first equation to solve for x and it came out as x=15.4
freckles
  • freckles
I'm not seeing how you obtained that
freckles
  • freckles
can you should your work so I can see how you first obtained y
freckles
  • freckles
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)
cutiecomittee123
  • cutiecomittee123
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
freckles
  • freckles
how did you get sqrt(5)?
cutiecomittee123
  • cutiecomittee123
After adding the equations I get -y^2=-5
freckles
  • freckles
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
cutiecomittee123
  • cutiecomittee123
I got square root 5 because I multiplied both sides by -1 and then took the square root to get y=sqrt5
cutiecomittee123
  • cutiecomittee123
So y^2 +4y=5
freckles
  • freckles
you can factor y^2+4y-5
cutiecomittee123
  • cutiecomittee123
Help me with that?
freckles
  • freckles
what two numbers multiply to be -5 and add up to be 4?
freckles
  • freckles
think -5 can be written as the following: -5(1) or 5(-1) -5+1=? 5+(-1)=?
cutiecomittee123
  • cutiecomittee123
-4 or 4
freckles
  • freckles
so which pair of numbers satisfy my question?
freckles
  • freckles
"what two numbers multiply to be -5 and add up to be 4?"
cutiecomittee123
  • cutiecomittee123
oh 5 and -1
freckles
  • freckles
y^2+4y-5=0 (y+5)(y-1)=0
cutiecomittee123
  • cutiecomittee123
(x+5) (y-1) = 0
freckles
  • freckles
so now you just set both factor equal to 0 y+5=0 or y-1=0
cutiecomittee123
  • cutiecomittee123
y=-5 and or y=1
freckles
  • freckles
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
cutiecomittee123
  • cutiecomittee123
x=2
freckles
  • freckles
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
freckles
  • freckles
actually you might end up with 4 solutions since x has a square on it
freckles
  • freckles
yep there is one more solution to x^2=4 I assume that is where you got x=2 from
cutiecomittee123
  • cutiecomittee123
x also =-2
freckles
  • freckles
so we have (2,-5) and (-2,-5) now we need to solve for x when y is 1
freckles
  • freckles
you can still use 2x^2+y^2=33
cutiecomittee123
  • cutiecomittee123
uh y=sqrt31
freckles
  • freckles
did you replace x with 1?
cutiecomittee123
  • cutiecomittee123
yeah
freckles
  • freckles
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
freckles
  • freckles
in other words we never said anything about x being 1
freckles
  • freckles
\[2x^2+y^2=33 \\ 2x^2+1^2=33 \\ 2x^2+1=33\] solve for x
cutiecomittee123
  • cutiecomittee123
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
freckles
  • freckles
why are you replacing x with 1? we never said anything about x being 1
cutiecomittee123
  • cutiecomittee123
Oh lol, I get x=4 after setting y=1
freckles
  • freckles
or also?
freckles
  • freckles
I assume you got that from x^2=16
freckles
  • freckles
but there is one more solution to that
cutiecomittee123
  • cutiecomittee123
x=-4
freckles
  • freckles
\[(2,-5); (-2,-5) ; (-4,1);(4,1)\]
cutiecomittee123
  • cutiecomittee123
sweet
cutiecomittee123
  • cutiecomittee123
okay now the question is which one of these is in quadrant 4 of the graph?
freckles
  • freckles
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)
freckles
  • freckles
|dw:1433531008630:dw|
cutiecomittee123
  • cutiecomittee123
(2,-5)
freckles
  • freckles
|dw:1433531053701:dw|
freckles
  • freckles
yep sounds great
cutiecomittee123
  • cutiecomittee123
Yes:)
NotTim
  • NotTim
lol this again
freckles
  • freckles
|dw:1433531133632:dw|
cutiecomittee123
  • cutiecomittee123
Can you help me with something else?:)
freckles
  • freckles
I can try
cutiecomittee123
  • cutiecomittee123
Okay its pretty simple what is the maximum number of solutions to this system x^2+4y=64 and x+y=5
freckles
  • freckles
well one is a parabola and one is a line
freckles
  • freckles
what is the max number of times a line can intersect a parabola?
cutiecomittee123
  • cutiecomittee123
2 times
freckles
  • freckles
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.
cutiecomittee123
  • cutiecomittee123
Awesome thanks
cutiecomittee123
  • cutiecomittee123
I have more but I think I am going to ask them in other questions so things dont get confusing
freckles
  • freckles
k
NotTim
  • NotTim
Wait so i was right in the other question? Dude convinced me otherwise. goddarnit.
NotTim
  • NotTim
wait no, here it was 2. i only got 1.
freckles
  • freckles
what other question?>

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