Question

system of equations, solve for x and y,
3x^2-2y and -3y^2-2x

Answer 1

where are the = signs

Answer 2

3x^2-2y=0 and -3y^2-2x=0

Answer 3

(0, 0)
(-2/3, 2/3)

Answer 4

LLort I found (0,0) but how did you get (-2/3,2/3)?

Answer 5

I graphed it

Answer 6

solve first one for y
y =(3x^2)/2
substitute it into 2nd equ
-3(9x^4/4) -2x = 0
multiply everything by 4
-27x^4 -8x = 0
-x(27x^3 +8) = 0
-x(3x+2)(9x^2 -6x +4) = 0
real solutions
x=0, x = -2/3
substitute in to find y
y =0, y = 2/3

Answer 7

dumbcow let me take a look at this, be a minute.

Answer 8

ok there are 4 solutions
(0,0)
(0,2/3)
(-2/3,0)
(-2/3,2/3)

Answer 9

How did 27x^3+8) become (3x+2)(9x^2 -6x +4)?

Answer 10

oh sorry, thats the sum of cubes formula
(a+b)^3 = (a+b)(a^2 -ab+b^2)
27x^3 = (3x)^3
8 = 2^3

Answer 11

*(a^3 +b^3)

Answer 12

I remember now. Thanks, the 2/3 is what i was looking for.