Chapter 4.6, Problem 31E

To determine

To solve: Inequality $\frac{1}{4a}+\frac{5}{8a}>\frac{1}{2}$

Answer to Problem 31E

  $0<a<\frac{7}{4}$

Explanation of Solution

Given information: $\frac{1}{4a}+\frac{5}{8a}>\frac{1}{2}$

Calculation:

  $\frac{1}{4a}+\frac{5}{8a}>\frac{1}{2}$

First make note of the domain

  $a\ne 0$

Multiply both sides by $8a$ .

  $\begin{array}{l}2+5=4a\\ a=\frac{7}{4}\end{array}$

Test points left and right of: $0,\frac{7}{4}$

  $\begin{array}{l}\text{Test:}\left(-1\right):\frac{1}{4\left(-1\right)}+\frac{5}{8\left(-1\right)}>\frac{1}{2}\\ \frac{7}{-8}>\frac{1}{2}\text{ False}\end{array}$

  $\begin{array}{l}\text{Test:}\left(1\right):\frac{1}{4}+\frac{5}{8}>\frac{1}{2}\\ \frac{7}{8}>\frac{1}{2}\text{ True}\end{array}$

  $\begin{array}{l}\text{Test:}\left(2\right):\frac{1}{4\left(2\right)}+\frac{5}{8\left(2\right)}>\frac{1}{2}\\ \frac{7}{16}>\frac{1}{2}\text{ False}\end{array}$

Solutions:

  $0<a<\frac{7}{4}$