Chapter A.9, Problem 87AYU

To determine

To solve: Each inequality. Express your answer using set notation or interval notation. Graph the solution set.

Answer to Problem 87AYU

$\left\{x|3 < x < \infty \right\}$ or $\left(3, \infty \right)$

Explanation of Solution

Given:

$0 < {\left(2x - 4\right)}^{- 1} < \frac{1}{2}$

Calculation:

$0 < {\left(2x - 4\right)}^{- 1} < \frac{1}{2}$

$0 < \frac{1}{2x - 4} < \frac{1}{2}$

$0 < \frac{1}{2\left(x - 2\right)} < \frac{1}{2}$

Multiply by 2

$0 < \frac{1}{x - 2} < 1$

$\frac{1}{x - 2} - 1 < 0$

$\frac{1 - x + 2}{x - 2} < 0$

$- x + 3 < 0$

$- x + 3 + x < 0 + x$

$3 < x$

$x > 3$

The solution set is $\left\{x|3 < x < \infty \right\}$ or using interval notation, all numbers in the interval $\left(3, \infty \right)$.