Can somebody show I how to solve simultaneouosly x^2 -3y=0 and -3x+3y^2=0 to get the point (0,0) and (1,1). I can't cancel anything out with those two equations so I'm stuck.

Answer 1

since y = x^2/3, use that in the 2nd equation to get
-3x + 3x^4/9 = 0
-3x + x^4/3 = 0
x^4 - 9x = 0
x = 0,∛9
so y = 0,∛3

Answer 2

That does not give the point (0,0) and (1,1). Something isn't right.

Answer 3

well, clearly (1,1) does not work!
x^3 - 3y = 0
??? 1^3 is not 3

Answer 4

If your system of equations is:

3 x² - 3 y = 0 and - 3 x + 3 y² = 0

then

3 x² - 3 y = 0

Divide both sides by 3

x² - y = 0

Add y to both sides

x² = y

y = x²

- 3 x + 3 y² = 0

Divide both sides by 3

- x + y² = 0

Replace y with x²

- x + ( x²)² = 0

- x + x⁴ = 0

x ( - 1 + x³ ) = 0

x ( x³ - 1 ) = 0

x ( x - 1 ) ( x² + x + 1 ) = 0

This equation is equal zero when:

x = 0 , x - 1 = 0 and x² + x + 1 = 0

x - 1 = 0

Add 1 to both sides

x = 1

The solutions of x² + x + 1 = 0 are:

x1 = - 1 / 2 + i √3 / 2 and x 2 = - 1 / 2 - i √3 / 2

So the real solutions are:

x = 0 and x =1

Replace this values in equation 3 x² - 3 y = 0

x = 0

3 ∙ 0² - 3 y = 0

- 3 y = 0

Divide both sides by - 3

y = 0

x = 1

3 ∙ 1² - 3 y = 0

3 - 3 y = 0

Add 3 y to both sides

3 = 3 y

3 y = 3

Divide both sides by 3

y = 1

This is correct only if your system of equations is 3 x² - 3 y = 0 and - 3 x + 3 y² = 0