Question

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

Answer 1

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

Answer 2

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

Answer 3

I did that but I got y=sqrt5

Answer 4

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

Answer 5

I'm not seeing how you obtained that

Answer 6

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

Answer 7

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)

Answer 8

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

Answer 9

how did you get sqrt(5)?

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

So y^2 +4y=5

Answer 14

you can factor y^2+4y-5

Answer 15

Help me with that?

Answer 16

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

Answer 17

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


-5+1=?

5+(-1)=?

Answer 18

-4 or 4

Answer 19

so which pair of numbers satisfy my question?

Answer 20

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

Answer 21

oh 5 and -1

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

y=-5 and or y=1

Answer 26

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

Answer 27

x=2

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

x also =-2

Answer 32

so we have (2,-5) and (-2,-5)

now we need to solve for x when y is 1

Answer 33

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

Answer 34

uh y=sqrt31

Answer 35

did you replace x with 1?

Answer 36

yeah

Answer 37

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

Answer 38

in other words we never said anything about x being 1

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

or also?

Answer 44

I assume you got that from x^2=16

Answer 45

but there is one more solution to that

Answer 46

x=-4

Answer 47

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

Answer 48

sweet

Answer 49

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

Answer 50

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)

Answer 51

(2,-5)

Answer 52

yep sounds great

Answer 53

Yes:)

Answer 54

lol this again