Lesson 2.7: Linear Inequalities and Absolute Value Inequalities

New Concept

This lesson covers solving linear and absolute value inequalities.
You'll apply algebraic properties, paying close attention to the rule for negative multipliers, and master expressing solution sets using interval notation, such as $[a, b)$ or $(-\infty, a]$.

What’s next

Next, you'll tackle worked examples on solving inequalities and then test your skills with a series of interactive practice cards and challenge problems.

Property

The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number.
Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be “equaled.”

Examples

• All real numbers greater than or equal to -3 is written in interval notation as $[-3, \infty)$.

• All real numbers between -4 and 5, including 5 but not -4, is written as $(-4, 5]$.

Property

Addition Property
If $a < b$, then $a + c < b + c$.

Multiplication Property
If $a < b$ and $c > 0$, then $ac < bc$.
If $a < b$ and $c < 0$, then $ac > bc$.

These properties also apply to $a \leq b$, $a > b$, and $a \geq b$.

Property

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.

Examples

• To solve $10 - 5x \geq 2x - 11$, combine x-terms to get $10 - 7x \geq -11$. Then isolate x: $-7x \geq -21$. Divide by -7 and flip the sign: $x \leq 3$, which is $(-\infty, 3]$.

• To solve $-x + 5 < \frac{1}{3}x + 1$, combine terms to get $4 < \frac{4}{3}x$. Multiply by $\frac{3}{4}$ to get $3 < x$. The interval is $(3, \infty)$.

Property

A compound inequality includes two inequalities in one statement. A statement such as $4 < x \leq 6$ means $4 < x$ and $x \leq 6$. 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.

Examples

• To solve $2 \leq 3x - 4 < 8$, add 4 to all three parts: $6 \leq 3x < 12$. Then divide all three parts by 3 to get $2 \leq x < 4$. The interval is $[2, 4)$.

• To solve $x+1 > 4x-5 > 3x-9$, split it into $x+1 > 4x-5$ and $4x-5 > 3x-9$. This gives $2 > x$ and $x > -4$. The solution is $(-4, 2)$.

Property

For an algebraic expression $X$, and $k > 0$, an absolute value inequality is an inequality of the form
$|X| < k$ is equivalent to $-k < X < k$
$|X| > k$ is equivalent to $X < -k$ or $X > k$
These statements also apply to $|X| \leq k$ and $|X| \geq k$.

Examples

• To describe all values $x$ within a distance of 2 from the number 7, write $|x - 7| \leq 2$. Solving $-2 \leq x - 7 \leq 2$ gives $5 \leq x \leq 9$, or $[5, 9]$.

• To solve $|x - 4| \leq 2$, set up the compound inequality $-2 \leq x - 4 \leq 2$. Adding 4 to all parts gives $2 \leq x \leq 6$, which is the interval $[2, 6]$.