Solve the System of Equations x^3+y=0 4x^2-y=0

$x^{3} + y = 0$ $4 x^{2} - y = 0$

Subtract $x^{3}$ from both sides of the equation.

$y = - x^{3}$

$4 x^{2} - y = 0$

Replace all occurrences of $y$ with $- x^{3}$ in each equation.

Replace all occurrences of $y$ in $4 x^{2} - y = 0$ with $- x^{3}$ .

$4 x^{2} - (- x^{3}) = 0$

$y = - x^{3}$

Simplify the left side.

Multiply $- (- x^{3})$ .

Multiply $- 1$ by $- 1$ .

$4 x^{2} + 1 x^{3} = 0$

$y = - x^{3}$

Multiply $x^{3}$ by $1$ .

$4 x^{2} + x^{3} = 0$

$y = - x^{3}$

$4 x^{2} + x^{3} = 0$

$y = - x^{3}$

$4 x^{2} + x^{3} = 0$

$y = - x^{3}$

$4 x^{2} + x^{3} = 0$

$y = - x^{3}$

Solve for $x$ in $4 x^{2} + x^{3} = 0$ .

Factor $x^{2}$ out of $4 x^{2} + x^{3}$ .

Factor $x^{2}$ out of $4 x^{2}$ .

$x^{2} \cdot 4 + x^{3} = 0$

$y = - x^{3}$

Factor $x^{2}$ out of $x^{3}$ .

$x^{2} \cdot 4 + x^{2} x = 0$

$y = - x^{3}$

Factor $x^{2}$ out of $x^{2} \cdot 4 + x^{2} x$ .

$x^{2} \cdot (4 + x) = 0$

$y = - x^{3}$

Multiply $x^{2}$ by $4 + x$ .

$x^{2} (4 + x) = 0$

$y = - x^{3}$

$x^{2} (4 + x) = 0$

$y = - x^{3}$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x^{2} = 0$

$4 + x = 0$

$y = - x^{3}$

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

$y = - x^{3}$

Solve $x^{2} = 0$ for $x$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

$y = - x^{3}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

$y = - x^{3}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

$y = - x^{3}$

Plus or minus $0$ is $0$ .

$x = 0$

$y = - x^{3}$

$x = 0$

$y = - x^{3}$

$x = 0$

$y = - x^{3}$

$x = 0$

$y = - x^{3}$

Set $4 + x$ equal to $0$ and solve for $x$ .

Set $4 + x$ equal to $0$ .

$4 + x = 0$

$y = - x^{3}$

Subtract $4$ from both sides of the equation.

$x = - 4$

$y = - x^{3}$

$x = - 4$

$y = - x^{3}$

The final solution is all the values that make $x^{2} (4 + x) = 0$ true.

$x = 0, - 4$

$y = - x^{3}$

$x = 0, - 4$

$y = - x^{3}$

Replace all occurrences of $x$ with $0$ in each equation.

Replace all occurrences of $x$ in $y = - x^{3}$ with $0$ .

$y = - {(0)}^{3}$

$x = 0$

Simplify the right side.

Simplify $- {(0)}^{3}$ .

Raising $0$ to any positive power yields $0$ .

$y = - 0$

$x = 0$

Multiply $- 1$ by $0$ .

$y = 0$

$x = 0$

$y = 0$

$x = 0$

$y = 0$

$x = 0$

$y = 0$

$x = 0$

Replace all occurrences of $x$ with $- 4$ in each equation.

Replace all occurrences of $x$ in $y = - x^{3}$ with $- 4$ .

$y = - {(- 4)}^{3}$

$x = - 4$

Simplify the right side.

Simplify $- {(- 4)}^{3}$ .

Raise $- 4$ to the power of $3$ .

$y = 64$

$x = - 4$

Multiply $- 1$ by $- 64$ .

$y = 64$

$x = - 4$

$y = 64$

$x = - 4$

$y = 64$

$x = - 4$

$y = 64$

$x = - 4$

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(0, 0)$

$(- 4, 64)$

The result can be shown in multiple forms.

Point Form:

$(0, 0), (- 4, 64)$

Equation Form:

$x = 0, y = 0$

$x = - 4, y = 64$

Solve the System of Equations x^3+y=0 4x^2-y=0

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