Solve the Absolute Value Inequality for x |2x-3|<=5

$| 2 x - 3 | \leq 5$

Write $| 2 x - 3 | \leq 5$ 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.

Add $3$ to 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 $1$ .

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

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

$2 x - 3 \leq 5$

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.

Add $3$ to 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 $1$ .

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

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

$- (2 x - 3) \leq 5$

Write as a piecewise.

${\begin{cases} 2 x - 3 \leq 5 & x \geq \frac{3}{2} \\ - (2 x - 3) \leq 5 & x < \frac{3}{2} \end{cases}$

Simplify $- (2 x - 3) \leq 5$ .

Apply the distributive property.

${\begin{cases} 2 x - 3 \leq 5 & x \geq \frac{3}{2} \\ - (2 x) - - 3 \leq 5 & x < \frac{3}{2} \end{cases}$

Multiply $2$ by $- 1$ .

${\begin{cases} 2 x - 3 \leq 5 & x \geq \frac{3}{2} \\ - 2 x - - 3 \leq 5 & x < \frac{3}{2} \end{cases}$

Multiply $- 1$ by $- 3$ .

${\begin{cases} 2 x - 3 \leq 5 & x \geq \frac{3}{2} \\ - 2 x + 3 \leq 5 & x < \frac{3}{2} \end{cases}$

Solve $2 x - 3 \leq 5$ when $x \geq \frac{3}{2}$ .

Solve $2 x - 3 \leq 5$ for $x$ .

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

Add $3$ to both sides of the inequality.

$2 x \leq 5 + 3$

Add $5$ and $3$ .

$2 x \leq 8$

Divide each term in $2 x \leq 8$ by $2$ and simplify.

Divide each term in $2 x \leq 8$ by $2$ .

$\frac{2 x}{2} \leq \frac{8}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \leq \frac{8}{2}$

Divide $x$ by $1$ .

$x \leq \frac{8}{2}$

Simplify the right side.

Divide $8$ by $2$ .

$x \leq 4$

Find the intersection of $x \leq 4$ and $x \geq \frac{3}{2}$ .

$\frac{3}{2} \leq x \leq 4$

Solve $- 2 x + 3 \leq 5$ when $x < \frac{3}{2}$ .

Solve $- 2 x + 3 \leq 5$ for $x$ .

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

Subtract $3$ from both sides of the inequality.

$- 2 x \leq 5 - 3$

Subtract $3$ from $5$ .

$- 2 x \leq 2$

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

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

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $2$ by $- 2$ .

$x \geq - 1$

Find the intersection of $x \geq - 1$ and $x < \frac{3}{2}$ .

$- 1 \leq x < \frac{3}{2}$

Find the union of the solutions.

$- 1 \leq x \leq 4$

The result can be shown in multiple forms.

Inequality Form:

$- 1 \leq x \leq 4$

Interval Notation:

$[- 1, 4]$

Solve the Absolute Value Inequality for x |2x-3|<=5

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