# 2.7 Linear inequalities and absolute value inequalities  (Page 3/11)

 Page 3 / 11

Solve the inequality and write the answer in interval notation: $\text{\hspace{0.17em}}-\frac{5}{6}x\le \frac{3}{4}+\frac{8}{3}x.$

$\left[-\frac{3}{14},\infty \right)$

## Understanding compound inequalities

A compound inequality    includes two inequalities in one statement. A statement such as $\text{\hspace{0.17em}}4 means $\text{\hspace{0.17em}}4 and $\text{\hspace{0.17em}}x\le 6.\text{\hspace{0.17em}}$ There are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time. We will illustrate both methods.

## Solving a compound inequality

Solve the compound inequality: $\text{\hspace{0.17em}}3\le 2x+2<6.$

The first method is to write two separate inequalities: $\text{\hspace{0.17em}}3\le 2x+2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}2x+2<6.\text{\hspace{0.17em}}$ We solve them independently.

$\begin{array}{lll}3\le 2x+2\hfill & \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\hfill & 2x+2<6\hfill \\ 1\le 2x\hfill & \hfill & \phantom{\rule{2em}{0ex}}2x<4\hfill \\ \frac{1}{2}\le x\hfill & \hfill & \phantom{\rule{2em}{0ex}}x<2\hfill \end{array}$

Then, we can rewrite the solution as a compound inequality, the same way the problem began.

$\frac{1}{2}\le x<2$

In interval notation, the solution is written as $\text{\hspace{0.17em}}\left[\frac{1}{2},2\right).$

The second method is to leave the compound inequality intact, and perform solving procedures on the three parts at the same time.

We get the same solution: $\text{\hspace{0.17em}}\left[\frac{1}{2},2\right).$

Solve the compound inequality: $\text{\hspace{0.17em}}4<2x-8\le 10.$

$6

## Solving a compound inequality with the variable in all three parts

Solve the compound inequality with variables in all three parts: $\text{\hspace{0.17em}}3+x>7x-2>5x-10.$

Let's try the first method. Write two inequalities :

$\begin{array}{lll}3+x>7x-2\hfill & \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\hfill & 7x-2>5x-10\hfill \\ \phantom{\rule{1.2em}{0ex}}3>6x-2\hfill & \hfill & 2x-2>-10\hfill \\ \phantom{\rule{1.2em}{0ex}}5>6x\hfill & \hfill & \phantom{\rule{1.8em}{0ex}}2x>-8\hfill \\ \phantom{\rule{1.2em}{0ex}}\frac{5}{6}>x\hfill & \hfill & \phantom{\rule{2.5em}{0ex}}x>-4\hfill \\ \phantom{\rule{1.3em}{0ex}}x<\frac{5}{6}\hfill & \hfill & \phantom{\rule{1.5em}{0ex}}-4

The solution set is $\text{\hspace{0.17em}}-4 or in interval notation $\text{\hspace{0.17em}}\left(-4,\frac{5}{6}\right).\text{\hspace{0.17em}}$ Notice that when we write the solution in interval notation, the smaller number comes first. We read intervals from left to right, as they appear on a number line. See [link] .

Solve the compound inequality: $\text{\hspace{0.17em}}3y<4-5y<5+3y.$

$\left(-\frac{1}{8},\frac{1}{2}\right)$

## Solving absolute value inequalities

As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at $\text{\hspace{0.17em}}\left(-x,0\right)\text{\hspace{0.17em}}$ has an absolute value of $\text{\hspace{0.17em}}x,$ as it is x units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero.

An absolute value inequality is an equation of the form

Where A , and sometimes B , represents an algebraic expression dependent on a variable x. Solving the inequality means finding the set of all $\text{\hspace{0.17em}}x$ - values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.

There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.

Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of$600. We can solve algebraically for the set of x- values such that the distance between $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and 600 is less than 200. We represent the distance between $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and 600 as $\text{\hspace{0.17em}}|x-600|,$ and therefore, $\text{\hspace{0.17em}}|x-600|\le 200\text{\hspace{0.17em}}$ or

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