What is the answer to 3yx^2-2yx=0,x^3-x^2=0 ?

The answer to 3yx^2-2yx=0,x^3-x^2=0 is y=0,x=0,x= 2/3

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3yx^2-2yx=0,x^3-x^2=0

3 y x^{2} - 2 y x = 0, x^{3} - x^{2} = 0

y = 0, x = 0, x = \frac{2}{3}

Decimal Notation

y = 0, x = 0, x = 0.66666 \dots

Solution steps

3 y x^{2} - 2 y x = 0

Factor

3 y x^{2} - 2 y x : y x (3 x - 2)

y x (3 x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y = 0 or x = 0 or 3 x - 2 = 0

y = 0

x = 0

Solve

3 x - 2 = 0 : x = \frac{2}{3}

The solutions are

y = 0, x = 0, x = \frac{2}{3}

(- 15 3 - x) (x^{2} + 15) = 0

- 4 x^{3} y^{4} = 0, - 4 x^{4} y^{3} = 0

I n (x^{2}) = 0

(x^{2} + 1) (y^{2} + y) = 0

(x^{2} + 1) y = 0

What is the answer to 3yx^2-2yx=0,x^3-x^2=0 ?
The answer to 3yx^2-2yx=0,x^3-x^2=0 is y=0,x=0,x= 2/3