# 2.7 Linear inequalities and absolute value inequalities  (Page 2/11)

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## Using interval notation to express all real numbers less than or equal to a Or greater than or equal to b

Write the interval expressing all real numbers less than or equal to $\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}$ or greater than or equal to $\text{\hspace{0.17em}}1.$

We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins at $\text{\hspace{0.17em}}-\infty \text{\hspace{0.17em}}$ and ends at $\text{\hspace{0.17em}}-1,$ which is written as $\text{\hspace{0.17em}}\left(-\infty ,-1\right].$

The second interval must show all real numbers greater than or equal to $\text{\hspace{0.17em}}1,$ which is written as $\text{\hspace{0.17em}}\left[1,\infty \right).\text{\hspace{0.17em}}$ However, we want to combine these two sets. We accomplish this by inserting the union symbol, $\cup ,$ between the two intervals.

$\left(-\infty ,-1\right]\cup \left[1,\infty \right)$

Express all real numbers less than $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ or greater than or equal to 3 in interval notation.

$\left(-\infty ,-2\right)\cup \left[3,\infty \right)$

## Using the properties of inequalities

When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol.

## Properties of inequalities

These properties also apply to $\text{\hspace{0.17em}}a\le b,$ $a>b,$ and $\text{\hspace{0.17em}}a\ge b.$

Illustrate the addition property for inequalities by solving each of the following:

• (a) $x-15<4$
• (b) $6\ge x-1$
• (c) $x+7>9$

The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.

Solve: $\text{\hspace{0.17em}}3x-2<1.$

$x<1$

## Demonstrating the multiplication property

Illustrate the multiplication property for inequalities by solving each of the following:

1. $3x<6$
2. $-2x-1\ge 5$
3. $5-x>10$

1. $\begin{array}{l}\phantom{\rule{1.5em}{0ex}}3x<6\hfill \\ \frac{1}{3}\left(3x\right)<\left(6\right)\frac{1}{3}\hfill \\ \phantom{\rule{2em}{0ex}}x<2\hfill \end{array}$

Solve: $\text{\hspace{0.17em}}4x+7\ge 2x-3.$

$x\ge -5$

## Solving inequalities in one variable algebraically

As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.

## Solving an inequality algebraically

Solve the inequality: $\text{\hspace{0.17em}}13-7x\ge 10x-4.$

Solving this inequality is similar to solving an equation up until the last step.

The solution set is given by the interval $\text{\hspace{0.17em}}\left(-\infty ,1\right],$ or all real numbers less than and including 1.

Solve the inequality and write the answer using interval notation: $\text{\hspace{0.17em}}-x+4<\frac{1}{2}x+1.$

$\left(2,\infty \right)$

## Solving an inequality with fractions

Solve the following inequality and write the answer in interval notation: $\text{\hspace{0.17em}}-\frac{3}{4}x\ge -\frac{5}{8}+\frac{2}{3}x.$

We begin solving in the same way we do when solving an equation.

The solution set is the interval $\text{\hspace{0.17em}}\left(-\infty ,\frac{15}{34}\right].$

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