Solve for x |2x-3|-5>=1

$| 2 x - 3 | - 5 \geq 1$

Write $| 2 x - 3 | - 5 \geq 1$ 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.

$2 x - 3 - 5 \geq 1$

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 $- 1$ .

$- (2 x - 3) - 5 \geq 1$

Write as a piecewise.

${\begin{cases} 2 x - 3 - 5 \geq 1 & x \geq \frac{3}{2} \\ - (2 x - 3) - 5 \geq 1 & x < \frac{3}{2} \end{cases}$

Subtract $5$ from $- 3$ .

${\begin{cases} 2 x - 8 \geq 1 & x \geq \frac{3}{2} \\ - (2 x - 3) - 5 \geq 1 & x < \frac{3}{2} \end{cases}$

Simplify $- (2 x - 3) - 5 \geq 1$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} 2 x - 8 \geq 1 & x \geq \frac{3}{2} \\ - (2 x) - - 3 - 5 \geq 1 & x < \frac{3}{2} \end{cases}$

Multiply $2$ by $- 1$ .

${\begin{cases} 2 x - 8 \geq 1 & x \geq \frac{3}{2} \\ - 2 x - - 3 - 5 \geq 1 & x < \frac{3}{2} \end{cases}$

Multiply $- 1$ by $- 3$ .

${\begin{cases} 2 x - 8 \geq 1 & x \geq \frac{3}{2} \\ - 2 x + 3 - 5 \geq 1 & x < \frac{3}{2} \end{cases}$

Subtract $5$ from $3$ .

${\begin{cases} 2 x - 8 \geq 1 & x \geq \frac{3}{2} \\ - 2 x - 2 \geq 1 & x < \frac{3}{2} \end{cases}$

Solve $2 x - 8 \geq 1$ for $x$ .

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

Add $8$ to both sides of the inequality.

$2 x \geq 1 + 8$

Add $1$ and $8$ .

$2 x \geq 9$

Divide each term in $2 x \geq 9$ by $2$ and simplify.

Divide each term in $2 x \geq 9$ by $2$ .

$\frac{2 x}{2} \geq \frac{9}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{9}{2}$

Divide $x$ by $1$ .

$x \geq \frac{9}{2}$

Solve $- 2 x - 2 \geq 1$ for $x$ .

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

Add $2$ to both sides of the inequality.

$- 2 x \geq 1 + 2$

Add $1$ and $2$ .

$- 2 x \geq 3$

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

Divide each term in $- 2 x \geq 3$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Find the union of the solutions.

$x \leq - \frac{3}{2}$ or $x \geq \frac{9}{2}$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - \frac{3}{2} or x \geq \frac{9}{2}$

Interval Notation:

$(- \infty, - \frac{3}{2}] \cup [\frac{9}{2}, \infty)$

Solve for x |2x-3|-5>=1

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