Question

Darn Systems of equations...

Answer 1

\[x ^{2}+x^2=9 and x^2 + (y-2)^2 =1\]

Answer 2

there's gonna be two answers

Answer 3

i think

Answer 4

is the first one \(y^2+x^2\) ?

Answer 5

no

Answer 6

oh it is

Answer 7

i just checked

Answer 8

@abb0t ?

Answer 9

I don't know how to do this. Let me find you a math prodigy to help you.

Answer 10

it is y^2 + x^2

Answer 11

intersection of 2 circles, there will be 2 points.
from 1st, you get
x^2 = 9-y^2
substitute this in 2nd equation to get a quadratic in y

Answer 12

i never thought of it as the intersection of 2 circles...im not sure how u got that though @gohangoku58

Answer 13

x^2+y^2=9
subtracting y^2 from both sides,
x^2 +y^2-y^2 = 9-y^2
x^2+0=9-y^2
x^2 =9-y^2
thats how...

Answer 14

you could just graph them both...

Answer 15

@gohangoku58 okay i see. do u substitute left or right side of the equal sign in the second equation?

Answer 16

so, for 2ns equation, u have
9-y^2 + (y-2)^2=1
can you simplify this ?
you'll get a simple linear equation.

Answer 17

expand (y-2)^2

Answer 18

kk

Answer 19

so \[2y^2 - 4y + 12\] ?

Answer 20

=0

Answer 21

and then i solve for y with the quadratic formula?

Answer 22

doesn't y^2 getting cancelled ?

Answer 23

**isn't

Answer 24

dang of course

Answer 25

so y=-3?

Answer 26

yes :)

Answer 27

kewl

Answer 28

so what am i supposed to do with this information?

Answer 29

do i plug y into an original equation?

Answer 30

find x
y= -3
put this in x^2+y^2 =9
yes!

Answer 31

so x=0?

Answer 32

wait!
i am gettting y= 3 !

Answer 33

-4y+12 = 0
4y=12
y=3

Answer 34

then, yes, x=0

Answer 35

kk

Answer 36

i forgot to to have the -'s cancel out..

Answer 37

so, the solution is 0,3

Answer 38

only.

Answer 39

kk, thx!

Answer 40

welcome ^_^