Solve for y 3y^2+y=0

$3 y^{2} + y = 0$

Factor $y$ out of $3 y^{2} + y$ .

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

$y (3 y) + y = 0$

Raise $y$ to the power of $1$ .

$y (3 y) + y = 0$

Factor $y$ out of $y^{1}$ .

$y (3 y) + y \cdot 1 = 0$

Factor $y$ out of $y (3 y) + y \cdot 1$ .

$y (3 y + 1) = 0$

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

$y = 0$

$3 y + 1 = 0$

Set $y$ equal to $0$ .

$y = 0$

Set $3 y + 1$ equal to $0$ and solve for $y$ .

Set $3 y + 1$ equal to $0$ .

$3 y + 1 = 0$

Solve $3 y + 1 = 0$ for $y$ .

Subtract $1$ from both sides of the equation.

$3 y = - 1$

Divide each term in $3 y = - 1$ by $3$ and simplify.

Divide each term in $3 y = - 1$ by $3$ .

$\frac{3 y}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{- 1}{3}$

Divide $y$ by $1$ .

$y = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{1}{3}$

The final solution is all the values that make $y (3 y + 1) = 0$ true.

$y = 0, - \frac{1}{3}$

The result can be shown in multiple forms.

Exact Form:

$y = 0, - \frac{1}{3}$

Decimal Form:

$y = 0, - 0.\overline{3}$