MA001: College Algebra

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

Addition Property

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

Multiplication Property

If $a < b$ and $c > 0$, then $a c < b c$.

If $a < b$ and $c < 0$, then $a c > b c$.

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

EXAMPLE 3

Demonstrating the Addition Property

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

1. (a) $x-15 < 4$

2. (b) $6 \geq x-1$

3. (c) $x+7 > 9$

Solution

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.

(a)

$\begin{array}{cl} x-15 < 4 \\ x-15+15 < 4+15 &\quad \text { Add } 15 \text { to both sides. } \\ x < 19 \end{array}$

(b)

$\begin{array}{cl} 6 \geq x-1 \\ 6+1 \geq x-1+1 & \qquad \text{Add 1 to both sides.}\\ 7 \geq x \end{array}$

(c)

$\begin{array}{cl} x+7 > 9 \\ x+7-7 > 9-7 & \qquad \text{Subtract 7 from both sides.}\\ x > 2 \end{array}$

TRY IT #3

Solve: $3 x-2 < 1$.

EXAMPLE 4

Demonstrating the Multiplication Property

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

1. (a) $3 x < 6$

2. (b) $-2 x-1 \geq 5$

3. (c) $5-x > 10$

Solution

(a)

$\begin{align} \begin{gathered} 3 x < 6 \\ \frac{1}{3}(3 x) < (6) \frac{1}{3} \\ x < 2 \end{gathered} \end{align}$

(b)

$\begin{array}{cl}-2 x-1 \geq 5 & \\ -2 x \geq 6 & \text { Multiply by }-\frac{1}{2} \\ \left(-\frac{1}{2}\right)(-2 x) \geq(6)\left(-\frac{1}{2}\right) & \text { Reverse the inequality. } \\ x \leq-3 & \end{array}$

(c)

$\begin{aligned} 5-x & > 10 & & \\-x & > 5 & & \\(-1)(-x) & > (5)(-1) & & \text { Multiply by }-1 \\ x & < -5 & & \text { Reverse the inequality } \end{aligned}$

TRY IT #4

Solve: $4 x+7 \geq 2 x-3$

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.

EXAMPLE 5

Solving an Inequality Algebraically

Solve the inequality: $13-7 x \geq 10 x-4$.

Solution

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

$\begin{array}{cl} 13-7 x \geq 10 x-4 & \\ 13-17 x \geq-4 & \text { Move variable terms to one side of the inequality. } \\ -17 x \geq-17 & \text { Isolate the variable term. } \\ x \leq 1 & \text { Dividing both sides by }-17 \text { reverses the inequality. } \end{array}$

The solution set is given by the interval $(-\infty, 1]$, or all real numbers less than and including $1$.

TRY IT #5

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

EXAMPLE 6

Solving an Inequality with Fractions

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

Solution

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

$\begin{aligned} -\frac{3}{4} x & \geq-\frac{5}{8}+\frac{2}{3} x & & \\ -\frac{3}{4} x-\frac{2}{3} x & \geq-\frac{5}{8} & & \text { Put variable terms on one side. } \\ -\frac{9}{12} x-\frac{8}{12} x & \geq-\frac{5}{8} & & \text { Write fractions with common denominator. } \\ -\frac{17}{12} x & \geq-\frac{5}{8} & \\ x & \leq-\frac{5}{8}\left(-\frac{12}{17}\right) & & \text { Multiplying by a negative number reverses the inequality. } \\ x & \leq \frac{15}{34} & & \end{aligned}$

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

TRY IT #6

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

Understanding Compound Inequalities

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. We will illustrate both methods.

EXAMPLE 7

Solving a Compound Inequality

Solve the compound inequality: $3 \leq 2 x+2 < 6$.

Solution

The first method is to write two separate inequalities: $3 \leq 2 x+2$ and $2 x+2 < 6$. We solve them independently.

$3 \leq 2 x+2 \quad$ and $\quad 2 x+2 < 6$

$1 \leq 2 x \quad 2 x < 4$

$\frac{1}{2} \leq x \quad x < 2$

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

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

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

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

$\begin{align} 3 \leq 2 x+2 < 6 \\ 1 \leq 2 x < 4 & \qquad \qquad \text{Isolate the variable term, and subtract 2 from all three parts.} \\ \frac{1}{2} \leq x < 2 & \qquad \qquad \text{Divide through all three parts by 2}.\end{align}$

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

TRY IT #7

Solve the compound inequality: $4 < 2 x-8 \leq 10$.

EXAMPLE 8

Solving a Compound Inequality with the Variable in All Three Parts

Solve the compound inequality with variables in all three parts: $3+x > 7 x-2 > 5 x-10$.

Solution

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

 $\begin{array}{ccr} 3+x > 7 x-2 & \qquad \quad \text { and }& \qquad 7 x-2 > 5 x-10\\ 3 > 6 x-2 & & 2 x-2 > -10\\ 5 > 6 x & & 2 x > -8\\ \frac{5}{6} > x & & x > -4\\ x < \frac{5}{6} & & -4 < x \end{array}$

The solution set is $-4 < x < \frac{5}{6}$ or in interval notation $\left(-4, \frac{5}{6}\right)$. 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 Figure 3.

Figure 3

TRY IT #8

Solve the compound inequality: $3y < 4-5y < 5+3y$.

Source: Rice University, https://openstax.org/books/college-algebra/pages/2-7-linear-inequalities-and-absolute-value-inequalities
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