Question

can someone please help me with this system of equations

Answer 1

its confusing

Answer 2

you are trying to get the solutions already listed?

Answer 3

yeah i just want to know how they got them

Answer 4

ok well I guess it is play time
I want to start out dividing the top two equations for 2 first
\[x+2=\lambda x \\ y-2=\lambda y\]
now I'm going to solve both of these for x and y in terms of lambda

Answer 5

\[x+2= \lambda x \\ 2=\lambda x -x \\ 2=(\lambda-1) x \\ x=\frac{2}{\lambda-1}\]
you should be able to solve the other equation for y in terms of lambda

Answer 6

what we will then do is replace the x and y in the x^2+y^2=9 with the expressions in terms of lambda given us a one variable equation

Answer 7

alright

Answer 8

so you solve \[y-2=\lambda y \text{ for } y\]

Answer 9

and let me know what you get

Answer 10

just after that one step

Answer 11

okay im getting \[y=\frac{ -2 }{ \lambda -1 }\]

Answer 12

sounds awesome
ok let's say what we have right now

Answer 13

\[x=\frac{2}{\lambda-1} \\ y=\frac{-2}{\lambda -1} \\ x^2+y^2=9\]

Answer 14

these are the equations we have now

Answer 15

we are going to plug the first two equations into the last equation there

Answer 16

\[(\frac{2}{\lambda-1})^2+(\frac{-2}{\lambda-1})^2=9 \text{ this equation is the same as } \\ (\frac{2}{\lambda-1})^2+(\frac{2}{\lambda-1})^2=9\]
do you know why?

Answer 17

because you get 4 on both

Answer 18

2^2 is the same as (-2)^2

or a^2 is the same as (-a)^2
since (-1)^2=1^2=1

Answer 19

cool so we have like terms then on the left hand side

Answer 20

\[2(\frac{2}{\lambda -1})^2=9\]

Answer 21

\[(\frac{2}{\lambda-1})^2=\frac{9}{2}\]
now take square root of both sides

Answer 22

you will have two equations from this

Answer 23

wait how you get the 2?

Answer 24

like terms

Answer 25

example u+u=2u

Answer 26

oh okay

Answer 27

\[\frac{2}{\lambda -1}=\sqrt{\frac{9}{2}} \text{ or } \frac{2}{\lambda-1}=-\sqrt{\frac{9}{2}}\]

Answer 28

instead of making this overly complicated remember what x and y equaled

Answer 29

we do not have to solve for lambda actually

Answer 30

stop right here and remember that x is 2/(lambda-1)
and y is -2/(lambda-1)

Answer 31

okay got it

Answer 32

\[\text{ the first equation says } \frac{2}{\lambda-1}=\sqrt{\frac{9}{2}} \\ x=\sqrt{\frac{9}{2}} \\ \text{ you could also multiply both sides by -1 } \\ \\ \text{ the second equation says } \frac{2}{\lambda-1}=-\sqrt{\frac{9}{2}} \\ x=-\sqrt{\frac{9}{2}}\]

Answer 33

wouldn't it be y=-sqrt(9/2) since y is the negative one?

Answer 34

or you could do this to both equations multiply both 1st and 2nd by -1
\[\frac{-2}{\lambda-1}=-\sqrt{\frac{9}{2}} \\ y=-\sqrt{\frac{9}{2}} \text{ from first equation } \\ \frac{-2}{\lambda-1}=\sqrt{\frac{9}{2}} \\ y=\sqrt{\frac{9}{2}} \text{ from second equation }\]

Answer 35

so when x is sqrt(9/2) y is -sqrt(9/2)
and when x is -sqrt(9/2) y is sqrt(9/2)

Answer 36

oh okay

Answer 37

so you see how I grouped the x and y from the first equation to get one ordered pair

and I grouped the x and y from the second equation together to get the other ordered pair

Answer 38

but the finding of y in that way is kinda unnecessary
we already knew that y is the opposite value of x due to the lambda expressions we wrote for both x and y

Answer 39

oh wow I just seen a neater and way more organized way to look at this

Answer 40

what is it?

Answer 41

let's make this a tiny more simpler

Answer 42

okay

Answer 43

we are still going to solve those two equations we had for x and y in terms of lambda

Answer 44

alright

Answer 45

\[x=\frac{2}{\lambda-1} \text{ and } y=\frac{-2}{\lambda-1}=-x\]

Answer 46

you see that y=-x right?

Answer 47

yes

Answer 48

assuming lambda is not 1

Answer 49

\[x^2+y^2=9 \\ \text{ since } y=-x \\ \text{ we have } \\ x^2+(-x)^2=9 \\ x^2+x^2=9 \\ 2x^2=9 \\ x^2=\frac{9}{2}\]

Answer 50

\[x=\sqrt{\frac{9}{2}} \text{ or } x=-\sqrt{\frac{9}{2}} \]

Answer 51

and remember y=-x

Answer 52

that is a much simpler way to do it

Answer 53

\[(x=\sqrt{\frac{9}{2}}, y=-x=-\sqrt{\frac{9}{2}})\]
you do something similar for the other pair

Answer 54

im looking at the problem right now and it showed another solution (-2,2). did they get that by solving for lambda?

Answer 55

that is weird

Answer 56

(-2)^2+(2)^2 isn't 9

Answer 57

oh lol nvm those are just the critical points

Answer 58

so the first equation would be \[(\frac{ 3 }{ \sqrt{2} },\frac{ -3 }{ \sqrt{2} }) \] right?

Answer 59

i mean not the first equation the first solution

Answer 60

yes

Answer 61

wouldn't it be easier if we start by eliminating the \(\lambda\) first by dividing first two equations ?

Answer 62

im not sure

Answer 63

i still dont get how you got the x to be negative for the second solution

Answer 64

do you know how to solve x^2+x^2=9?

Answer 65

you should get two values for x

Answer 66

similar way you get two values for x when you solve x^2=25
x=5 or x=-5
since 5^2=25 and (-5)^2=25

Answer 67

okay but we only got one value for x

Answer 68

no

Answer 69

we got two

Answer 70

we didnt?

Answer 71

\[x=\sqrt{\frac{9}{2}} \text{ or } x=-\sqrt{\frac{9}{2}} \]

Answer 72

\[x^2+y^2=9 \\ \text{ since } y=-x \\ \text{ we have } \\ x^2+(-x)^2=9 \\ x^2+x^2=9 \\ 2x^2=9 \\ x^2=\frac{9}{2} \]
this was the work that came before it

Answer 73

oh my bad didnt see that part

Answer 74

do you think it could be possible to do it the way ganeshie said

Answer 75

I eliminated the lambda
I don't know if he meant to eliminate lambda in a different way or not

Answer 76

let me think to see if I can think of a different more easier way that realizing y=-x
and plugging that into the last equation

Answer 77

its okay you dont have to i was just asking if it was possible

Answer 78

hmmm... I guess instead we could multiply -y to first equation
and x to the second equation giving us

\[-2xy-4y=-2 \lambda xy\]
and
\[2xy-4x=2 \lambda xy\]

now add the equations giving you
-4y-4x=0
-y-x=0
or
-x=y
y=-x

don't know if this way is more simpler than solving the first equation for x and the second one for y and then seeing that y=-x or not

to me that is opinion

Answer 79

i think either way you go you have to solve x^2+x^2=9

Answer 80

and then use y=-x to find the 2nd component of both ordered pairs

Answer 81

yeah your way looks a lot more simpler

Answer 82

let me know if you are still fighting with the answer

Answer 83

alright thanks