Solve the Inequality for x 9x>=3/2+2(6x+5)+2x

$9 x \geq \frac{3}{2} + 2 (6 x + 5) + 2 x$

Rewrite so $x$ is on the left side of the inequality.

$\frac{3}{2} + 2 (6 x + 5) + 2 x \leq 9 x$

Simplify $\frac{3}{2} + 2 (6 x + 5) + 2 x$ .

Find the common denominator.

Write $2 (6 x + 5)$ as a fraction with denominator $1$ .

$\frac{3}{2} + \frac{2 (6 x + 5)}{1} + 2 x \leq 9 x$

Multiply $\frac{2 (6 x + 5)}{1}$ by $\frac{2}{2}$ .

$\frac{3}{2} + \frac{2 (6 x + 5)}{1} \cdot \frac{2}{2} + 2 x \leq 9 x$

Multiply $\frac{2 (6 x + 5)}{1}$ by $\frac{2}{2}$ .

$\frac{3}{2} + \frac{2 (6 x + 5) \cdot 2}{2} + 2 x \leq 9 x$

Write $2 x$ as a fraction with denominator $1$ .

$\frac{3}{2} + \frac{2 (6 x + 5) \cdot 2}{2} + \frac{2 x}{1} \leq 9 x$

Multiply $\frac{2 x}{1}$ by $\frac{2}{2}$ .

$\frac{3}{2} + \frac{2 (6 x + 5) \cdot 2}{2} + \frac{2 x}{1} \cdot \frac{2}{2} \leq 9 x$

Multiply $\frac{2 x}{1}$ by $\frac{2}{2}$ .

$\frac{3}{2} + \frac{2 (6 x + 5) \cdot 2}{2} + \frac{2 x \cdot 2}{2} \leq 9 x$

Combine the numerators over the common denominator.

$\frac{3 + 2 (6 x + 5) \cdot 2 + 2 x \cdot 2}{2} \leq 9 x$

Simplify each term.

Apply the distributive property.

$\frac{3 + (2 (6 x) + 2 \cdot 5) \cdot 2 + 2 x \cdot 2}{2} \leq 9 x$

Multiply $6$ by $2$ .

$\frac{3 + (12 x + 2 \cdot 5) \cdot 2 + 2 x \cdot 2}{2} \leq 9 x$

Multiply $2$ by $5$ .

$\frac{3 + (12 x + 10) \cdot 2 + 2 x \cdot 2}{2} \leq 9 x$

Apply the distributive property.

$\frac{3 + 12 x \cdot 2 + 10 \cdot 2 + 2 x \cdot 2}{2} \leq 9 x$

Multiply $2$ by $12$ .

$\frac{3 + 24 x + 10 \cdot 2 + 2 x \cdot 2}{2} \leq 9 x$

Multiply $10$ by $2$ .

$\frac{3 + 24 x + 20 + 2 x \cdot 2}{2} \leq 9 x$

Multiply $2$ by $2$ .

$\frac{3 + 24 x + 20 + 4 x}{2} \leq 9 x$

Simplify by adding terms.

Add $3$ and $20$ .

$\frac{24 x + 23 + 4 x}{2} \leq 9 x$

Add $24 x$ and $4 x$ .

$\frac{28 x + 23}{2} \leq 9 x$

Move all terms containing $x$ to the left side of the inequality.

Subtract $9 x$ from both sides of the inequality.

$\frac{28 x + 23}{2} - 9 x \leq 0$

To write $- 9 x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{28 x + 23}{2} - 9 x \cdot \frac{2}{2} \leq 0$

Combine $- 9 x$ and $\frac{2}{2}$ .

$\frac{28 x + 23}{2} + \frac{- 9 x \cdot 2}{2} \leq 0$

Combine the numerators over the common denominator.

$\frac{28 x + 23 - 9 x \cdot 2}{2} \leq 0$

Simplify the numerator.

Multiply $2$ by $- 9$ .

$\frac{28 x + 23 - 18 x}{2} \leq 0$

Subtract $18 x$ from $28 x$ .

$\frac{10 x + 23}{2} \leq 0$

Multiply both sides by $2$ .

$\frac{10 x + 23}{2} \cdot 2 \leq 0 \cdot 2$

Simplify.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{10 x + 23}{2} \cdot 2 \leq 0 \cdot 2$

Rewrite the expression.

$10 x + 23 \leq 0 \cdot 2$

Simplify the right side.

Multiply $0$ by $2$ .

$10 x + 23 \leq 0$

Solve for $x$ .

Subtract $23$ from both sides of the inequality.

$10 x \leq - 23$

Divide each term in $10 x \leq - 23$ by $10$ and simplify.

Divide each term in $10 x \leq - 23$ by $10$ .

$\frac{10 x}{10} \leq \frac{- 23}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 x}{10} \leq \frac{- 23}{10}$

Divide $x$ by $1$ .

$x \leq \frac{- 23}{10}$

Simplify the right side.

Move the negative in front of the fraction.

$x \leq - \frac{23}{10}$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - \frac{23}{10}$

Interval Notation:

$(- \infty, - \frac{23}{10}]$

Solve the Inequality for x 9x>=3/2+2(6x+5)+2x

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