Solve the Inequality for y 3y^2-2y-3y(y-6)>=-2

$3 y^{2} - 2 y - 3 y (y - 6) \geq - 2$

Step 1

Simplify $3 y^{2} - 2 y - 3 y (y - 6)$ .

Simplify each term.

Apply the distributive property.

$3 y^{2} - 2 y - 3 y \cdot y - 3 y \cdot - 6 \geq - 2$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$3 y^{2} - 2 y - 3 (y \cdot y) - 3 y \cdot - 6 \geq - 2$

Multiply $y$ by $y$ .

$3 y^{2} - 2 y - 3 y^{2} - 3 y \cdot - 6 \geq - 2$

Multiply $- 6$ by $- 3$ .

$3 y^{2} - 2 y - 3 y^{2} + 18 y \geq - 2$

Simplify by adding terms.

Combine the opposite terms in $3 y^{2} - 2 y - 3 y^{2} + 18 y$ .

Subtract $3 y^{2}$ from $3 y^{2}$ .

$- 2 y + 0 + 18 y \geq - 2$

Add $- 2 y$ and $0$ .

$- 2 y + 18 y \geq - 2$

Add $- 2 y$ and $18 y$ .

$16 y \geq - 2$

Step 2

Divide each term in $16 y \geq - 2$ by $16$ and simplify.

Divide each term in $16 y \geq - 2$ by $16$ .

$\frac{16 y}{16} \geq \frac{- 2}{16}$

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

$\frac{16 y}{16} \geq \frac{- 2}{16}$

Divide $y$ by $1$ .

$y \geq \frac{- 2}{16}$

Simplify the right side.

Cancel the common factor of $- 2$ and $16$ .

Factor $2$ out of $- 2$ .

$y \geq \frac{2 (- 1)}{16}$

Cancel the common factors.

Factor $2$ out of $16$ .

$y \geq \frac{2 \cdot - 1}{2 \cdot 8}$

Cancel the common factor.

$y \geq \frac{2 \cdot - 1}{2 \cdot 8}$

Rewrite the expression.

$y \geq \frac{- 1}{8}$

Move the negative in front of the fraction.

$y \geq - \frac{1}{8}$

Step 3

The result can be shown in multiple forms.

Inequality Form:

$y \geq - \frac{1}{8}$

Interval Notation:

$[- \frac{1}{8}, \infty)$

Solve the Inequality for y 3y^2-2y-3y(y-6)>=-2

Related Questions

Support LCMGCF.com by disabling your adblocker.