What is the answer to 2x(y+1)^2=0,2x^2(y+1)=0 ?

The answer to 2x(y+1)^2=0,2x^2(y+1)=0 is x=0,y=-1

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2x(y+1)^2=0,2x^2(y+1)=0

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

x = 0, y = - 1

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or y + 1 = 0

x = 0

Solve

y + 1 = 0 : y = - 1

The solutions are

x = 0, y = - 1

(x - 4) (y + 6) = 0

(1 + z^{6}) (x) = 0

y 0 - (1 + x) y = 0

X^{2} + Y^{2} - X^{2} + 6 Y = 0

y, 6 x y - 12 x = 0.3 x^{2} + 3 y^{2} - 12 y

What is the answer to 2x(y+1)^2=0,2x^2(y+1)=0 ?
The answer to 2x(y+1)^2=0,2x^2(y+1)=0 is x=0,y=-1