Chapter A.9, Problem 130AYU

To determine

To find: Do you prefer to use inequality notation or interval notation to express the solution to an inequality? Give your reasons. Are there particular circumstances when you prefer one to the other? Cite examples.

Answer to Problem 130AYU

Using an inequality notation or inequality interval to express the solution of an inequality depends on the nature of the solution.

If there are infinite number of values of given variable in the inequality, that satisfy the given inequality, it’s always better to express the inequality notion like $>, <, \ge , \le$ as it shows an infinite range of solutions.

Example:

$\begin{array}{l}1 - 2x \le 3\\ 1 - 2x - 1 \le 3 - 1\\ - 2x \le 2\\ \frac{-2x}{-2} \ge \frac{2}{-2}\\ x \ge - 1\end{array}$

The solution set is $\left\{x|x \ge - 1\right\}$

If there are finite number of values of given variable in the inequality, that satisfy the given inequality, it’s always better to express the inequality interval.

Example:

$\begin{array}{l}4 \le 2x + 2 \le 10\\ 4 - 2 \le 2x + 2 - 2 \le 10 - 2\\ 2 \le 2x \le 8\\ \frac{2}{2} \le \frac{2x}{2} \le \frac{8}{2}\\ 1 \le x \le 4\end{array}$

The solution set is $\left\{x|1 \le x \le 4\right\}$ or using interval notation, all numbers in the interval $\left[1, 4\right]$

Explanation of Solution

Using an inequality notation or inequality interval to express the solution of an inequality depends on the nature of the solution.

If there are infinite number of values of given variable in the inequality, that satisfy the given inequality, it’s always better to express the inequality notion like $>, <, \ge , \le$ as it shows an infinite range of solutions.

Example:

$\begin{array}{l}1 - 2x \le 3\\ 1 - 2x - 1 \le 3 - 1\\ - 2x \le 2\\ \frac{-2x}{-2} \ge \frac{2}{-2}\\ x \ge - 1\end{array}$

The solution set is $\left\{x|x \ge - 1\right\}$

If there are finite number of values of given variable in the inequality, that satisfy the given inequality, it’s always better to express the inequality interval.

Example:

$\begin{array}{l}4 \le 2x + 2 \le 10\\ 4 - 2 \le 2x + 2 - 2 \le 10 - 2\\ 2 \le 2x \le 8\\ \frac{2}{2} \le \frac{2x}{2} \le \frac{8}{2}\\ 1 \le x \le 4\end{array}$

The solution set is $\left\{x|1 \le x \le 4\right\}$ or using interval notation, all numbers in the interval $\left[1, 4\right]$