Solve the Absolute Value Inequality for x |2x+3|>=19

$| 2 x + 3 | \geq 19$

Write $| 2 x + 3 | \geq 19$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$2 x + 3 \geq 0$

Solve the inequality.

Subtract $3$ from both sides of the inequality.

$2 x \geq - 3$

Divide each term in $2 x \geq - 3$ by $2$ and simplify.

Divide each term in $2 x \geq - 3$ by $2$ .

$\frac{2 x}{2} \geq \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{- 3}{2}$

Divide $x$ by $1$ .

$x \geq \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{3}{2}$

In the piece where $2 x + 3$ is non-negative, remove the absolute value.

$2 x + 3 \geq 19$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$2 x + 3 < 0$

Solve the inequality.

Subtract $3$ from both sides of the inequality.

$2 x < - 3$

Divide each term in $2 x < - 3$ by $2$ and simplify.

Divide each term in $2 x < - 3$ by $2$ .

$\frac{2 x}{2} < \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} < \frac{- 3}{2}$

Divide $x$ by $1$ .

$x < \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x < - \frac{3}{2}$

In the piece where $2 x + 3$ is negative, remove the absolute value and multiply by $- 1$ .

$- (2 x + 3) \geq 19$

Write as a piecewise.

${\begin{cases} 2 x + 3 \geq 19 & x \geq - \frac{3}{2} \\ - (2 x + 3) \geq 19 & x < - \frac{3}{2} \end{cases}$

Simplify $- (2 x + 3) \geq 19$ .

Apply the distributive property.

${\begin{cases} 2 x + 3 \geq 19 & x \geq - \frac{3}{2} \\ - (2 x) - 1 \cdot 3 \geq 19 & x < - \frac{3}{2} \end{cases}$

Multiply $2$ by $- 1$ .

${\begin{cases} 2 x + 3 \geq 19 & x \geq - \frac{3}{2} \\ - 2 x - 1 \cdot 3 \geq 19 & x < - \frac{3}{2} \end{cases}$

Multiply $- 1$ by $3$ .

${\begin{cases} 2 x + 3 \geq 19 & x \geq - \frac{3}{2} \\ - 2 x - 3 \geq 19 & x < - \frac{3}{2} \end{cases}$

Solve $2 x + 3 \geq 19$ for $x$ .

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

Subtract $3$ from both sides of the inequality.

$2 x \geq 19 - 3$

Subtract $3$ from $19$ .

$2 x \geq 16$

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

Divide each term in $2 x \geq 16$ by $2$ .

$\frac{2 x}{2} \geq \frac{16}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{16}{2}$

Divide $x$ by $1$ .

$x \geq \frac{16}{2}$

Simplify the right side.

Divide $16$ by $2$ .

$x \geq 8$

Solve $- 2 x - 3 \geq 19$ for $x$ .

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

Add $3$ to both sides of the inequality.

$- 2 x \geq 19 + 3$

Add $19$ and $3$ .

$- 2 x \geq 22$

Divide each term in $- 2 x \geq 22$ by $- 2$ and simplify.

Divide each term in $- 2 x \geq 22$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 x}{- 2} \leq \frac{22}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} \leq \frac{22}{- 2}$

Divide $x$ by $1$ .

$x \leq \frac{22}{- 2}$

Simplify the right side.

Divide $22$ by $- 2$ .

$x \leq - 11$

Find the union of the solutions.

$x \leq - 11$ or $x \geq 8$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 11 or x \geq 8$

Interval Notation:

$(- \infty, - 11] \cup [8, \infty)$

Solve the Absolute Value Inequality for x |2x+3|>=19

Related Questions

Support LCMGCF.com by disabling your adblocker.