Solve for x |2x+1|<13

$| 2 x + 1 | < 13$

Step 1

Write $| 2 x + 1 | < 13$ as a piecewise.

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

$2 x + 1 \geq 0$

Solve the inequality.

Subtract $1$ from both sides of the inequality.

$2 x \geq - 1$

Divide each term in $2 x \geq - 1$ by $2$ and simplify.

Divide each term in $2 x \geq - 1$ by $2$ .

$\frac{2 x}{2} \geq \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{- 1}{2}$

Divide $x$ by $1$ .

$x \geq \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{1}{2}$

In the piece where $2 x + 1$ is non-negative, remove the absolute value.

$2 x + 1 < 13$

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

$2 x + 1 < 0$

Solve the inequality.

Subtract $1$ from both sides of the inequality.

$2 x < - 1$

Divide each term in $2 x < - 1$ by $2$ and simplify.

Divide each term in $2 x < - 1$ by $2$ .

$\frac{2 x}{2} < \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} < \frac{- 1}{2}$

Divide $x$ by $1$ .

$x < \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x < - \frac{1}{2}$

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

$- (2 x + 1) < 13$

Write as a piecewise.

${\begin{cases} 2 x + 1 < 13 & x \geq - \frac{1}{2} \\ - (2 x + 1) < 13 & x < - \frac{1}{2} \end{cases}$

Simplify $- (2 x + 1) < 13$ .

Apply the distributive property.

${\begin{cases} 2 x + 1 < 13 & x \geq - \frac{1}{2} \\ - (2 x) - 1 \cdot 1 < 13 & x < - \frac{1}{2} \end{cases}$

Multiply $2$ by $- 1$ .

${\begin{cases} 2 x + 1 < 13 & x \geq - \frac{1}{2} \\ - 2 x - 1 \cdot 1 < 13 & x < - \frac{1}{2} \end{cases}$

Multiply $- 1$ by $1$ .

${\begin{cases} 2 x + 1 < 13 & x \geq - \frac{1}{2} \\ - 2 x - 1 < 13 & x < - \frac{1}{2} \end{cases}$

Step 2

Solve $2 x + 1 < 13$ when $x \geq - \frac{1}{2}$ .

Solve $2 x + 1 < 13$ for $x$ .

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

Subtract $1$ from both sides of the inequality.

$2 x < 13 - 1$

Subtract $1$ from $13$ .

$2 x < 12$

Divide each term in $2 x < 12$ by $2$ and simplify.

Divide each term in $2 x < 12$ by $2$ .

$\frac{2 x}{2} < \frac{12}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} < \frac{12}{2}$

Divide $x$ by $1$ .

$x < \frac{12}{2}$

Simplify the right side.

Divide $12$ by $2$ .

$x < 6$

Find the intersection of $x < 6$ and $x \geq - \frac{1}{2}$ .

$- \frac{1}{2} \leq x < 6$

Step 3

Solve $- 2 x - 1 < 13$ when $x < - \frac{1}{2}$ .

Solve $- 2 x - 1 < 13$ for $x$ .

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

Add $1$ to both sides of the inequality.

$- 2 x < 13 + 1$

Add $13$ and $1$ .

$- 2 x < 14$

Divide each term in $- 2 x < 14$ by $- 2$ and simplify.

Divide each term in $- 2 x < 14$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 x}{- 2} > \frac{14}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} > \frac{14}{- 2}$

Divide $x$ by $1$ .

$x > \frac{14}{- 2}$

Simplify the right side.

Divide $14$ by $- 2$ .

$x > - 7$

Find the intersection of $x > - 7$ and $x < - \frac{1}{2}$ .

$- 7 < x < - \frac{1}{2}$

Step 4

Find the union of the solutions.

$- 7 < x < 6$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$- 7 < x < 6$

Interval Notation:

$(- 7, 6)$

Step 6

Solve for x |2x+1|<13

Related Questions

Support LCMGCF.com by disabling your adblocker.