Question

critical points:

6x−1−5y2=0
10y2−10xy=0

Answer 1

I was right on my method once I remembered it. I don't blame you for doubting me though.

Answer 2

hi there.. i didnt understand what you did in the other topic.. so my question is re-formulated :)

Answer 3

could you please explain it more detailed so i can understand?

Answer 4

but what i did was that wrong? 10y(y-x) = 0 so y must be zero

Answer 5

yes that is ONE solution, but because this system is non-linear you will get three, watch:

Answer 6

oke.. i will wait for your explanation :)

Answer 7

sure

6x−1−5y2=0 (1)
10y2−10xy=0 (2)

solve (1) for x:

6x=5y^2+1
x=(5y^2+1)/6

sub into (2):

10y^2-(10/6)(5y^2+1)y=0
10y^2-25/3y^3-5y/3=0

factor out (-5y/3):

(-5y/3)(5y^2-6y+1)=0
(-5y/3)(5y-1)(y-1)=0
y={0,1/5,1}

now do what you did for each y separately:

for y=0:
6x-1=0->x=1/6

for y=1/5:
6x-1-5(1/5)^2=6x-1-1/5=0->x=1/5

for y=1
6x-1-5=0-->x=1

so the critical points are
(0,1/6) (1/5,1/5) (1,1)

Answer 8

It's always possible I made an algebraic mistake somewhere, so please check my work, but the method is kosher.

Answer 9

wooh oke.. i'm gonna read it now :)

Answer 10

ahhh i understand it now :)

thank you for your time and
explanation <3

Answer 11

thanks for refreshing my memory on this stuff :)

Answer 12

thank you :P