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

- cutiecomittee123

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

- Stacey Warren - Expert brainly.com

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- katieb

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- freckles

I would attempt to eliminate the x^2 term and solve for y first then go back and solve for x

- 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

I did that but I got y=sqrt5

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- cutiecomittee123

then I plugged it back into the first equation to solve for x and it came out as x=15.4

- freckles

I'm not seeing how you obtained that

- freckles

can you should your work so I can see how you first obtained y

- 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

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

how did you get sqrt(5)?

- cutiecomittee123

After adding the equations I get -y^2=-5

- 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

I got square root 5 because I multiplied both sides by -1 and then took the square root to get y=sqrt5

- cutiecomittee123

So y^2 +4y=5

- freckles

you can factor y^2+4y-5

- cutiecomittee123

Help me with that?

- freckles

what two numbers multiply to be -5 and add up to be 4?

- freckles

think
-5 can be written as the following:
-5(1)
or
5(-1)
-5+1=?
5+(-1)=?

- cutiecomittee123

-4 or 4

- freckles

so which pair of numbers satisfy my question?

- freckles

"what two numbers multiply to be -5 and add up to be 4?"

- cutiecomittee123

oh 5 and -1

- freckles

y^2+4y-5=0
(y+5)(y-1)=0

- cutiecomittee123

(x+5) (y-1) = 0

- freckles

so now you just set both factor equal to 0
y+5=0 or y-1=0

- cutiecomittee123

y=-5 and or y=1

- 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

x=2

- 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

actually you might end up with 4 solutions since x has a square on it

- freckles

yep there is one more solution to x^2=4
I assume that is where you got x=2 from

- cutiecomittee123

x also =-2

- freckles

so we have (2,-5) and (-2,-5)
now we need to solve for x when y is 1

- freckles

you can still use 2x^2+y^2=33

- cutiecomittee123

uh y=sqrt31

- freckles

did you replace x with 1?

- cutiecomittee123

yeah

- 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

in other words we never said anything about x being 1

- freckles

\[2x^2+y^2=33 \\ 2x^2+1^2=33 \\ 2x^2+1=33\]
solve for x

- 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

why are you replacing x with 1?
we never said anything about x being 1

- cutiecomittee123

Oh lol, I get x=4 after setting y=1

- freckles

or also?

- freckles

I assume you got that from x^2=16

- freckles

but there is one more solution to that

- cutiecomittee123

x=-4

- freckles

\[(2,-5); (-2,-5) ; (-4,1);(4,1)\]

- cutiecomittee123

sweet

- cutiecomittee123

okay now the question is which one of these is in quadrant 4 of the graph?

- 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

|dw:1433531008630:dw|

- cutiecomittee123

(2,-5)

- freckles

|dw:1433531053701:dw|

- freckles

yep sounds great

- cutiecomittee123

Yes:)

- NotTim

lol this again

- freckles

|dw:1433531133632:dw|

- cutiecomittee123

Can you help me with something else?:)

- freckles

I can try

- cutiecomittee123

Okay its pretty simple
what is the maximum number of solutions to this system
x^2+4y=64
and x+y=5

- freckles

well one is a parabola and one is a line

- freckles

what is the max number of times a line can intersect a parabola?

- cutiecomittee123

2 times

- 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

Awesome thanks

- cutiecomittee123

I have more but I think I am going to ask them in other questions so things dont get confusing

- freckles

k

- NotTim

Wait so i was right in the other question? Dude convinced me otherwise. goddarnit.

- NotTim

wait no, here it was 2. i only got 1.

- freckles

what other question?>

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