Solve the Inequality for x 6(2x+1/2)+3<=-3

$6 (2 x + \frac{1}{2}) + 3 \leq - 3$

Simplify $6 (2 x + \frac{1}{2}) + 3$ .

Simplify each term.

Apply the distributive property.

$6 (2 x) + 6 (\frac{1}{2}) + 3 \leq - 3$

Multiply $2$ by $6$ .

$12 x + 6 (\frac{1}{2}) + 3 \leq - 3$

Cancel the common factor of $2$ .

Factor $2$ out of $6$ .

$12 x + 2 (3) \frac{1}{2} + 3 \leq - 3$

Cancel the common factor.

$12 x + 2 \cdot 3 \frac{1}{2} + 3 \leq - 3$

Rewrite the expression.

$12 x + 3 + 3 \leq - 3$

Add $3$ and $3$ .

$12 x + 6 \leq - 3$

Move all terms not containing $x$ to the right side of the inequality.

Subtract $6$ from both sides of the inequality.

$12 x \leq - 3 - 6$

Subtract $6$ from $- 3$ .

$12 x \leq - 9$

Divide each term in $12 x \leq - 9$ by $12$ and simplify.

Divide each term in $12 x \leq - 9$ by $12$ .

$\frac{12 x}{12} \leq \frac{- 9}{12}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 x}{12} \leq \frac{- 9}{12}$

Divide $x$ by $1$ .

$x \leq \frac{- 9}{12}$

Simplify the right side.

Cancel the common factor of $- 9$ and $12$ .

Factor $3$ out of $- 9$ .

$x \leq \frac{3 (- 3)}{12}$

Cancel the common factors.

Factor $3$ out of $12$ .

$x \leq \frac{3 \cdot - 3}{3 \cdot 4}$

Cancel the common factor.

$x \leq \frac{3 \cdot - 3}{3 \cdot 4}$

Rewrite the expression.

$x \leq \frac{- 3}{4}$

Move the negative in front of the fraction.

$x \leq - \frac{3}{4}$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - \frac{3}{4}$

Interval Notation:

$(- \infty, - \frac{3}{4}]$

Solve the Inequality for x 6(2x+1/2)+3<=-3

Related Questions

Support LCMGCF.com by disabling your adblocker.