Inequalities

### $=, \: \gt, \: \lt, \: \ge, \: \le$

Sometimes we're less interested in solving an equality than an inequality. Here are the possibilities and symbols:

Expression Meaning
$a = b$ a is equal to b
$a \gt b$ a is greater than b
$a \lt b$ a is less than b
$a \ge b$ a is greater than or equal to b
$a \le b$ a is less than or equal to b
$a \ne b$ a is not equal to b

With one important exception, we use the same algebra techniques or "moves" to solve inequalities as equations. Let's get that exception out of the way first.

The exception: multiplying or dividing by a negative number

For example, it is obviously true that   $3 \gt 2.$ Now let's take that inequality and, following the rules for an equation, let's multiply both sides by -1.

\begin{align} 3 \cdot (-1) &\gt 2 \cdot (-1)\\[5pt] -3 &\gt -2, \end{align}

which is not true.

When working with inequalities, if we divide or multiply by a negative number, we have to change the direction of the > or < sign. The same goes for ≥ and ≤ signs.

#### Algebra with inequalities

The algebra of equations and inequalities is the same, except that when we multiply or divide both sides of an inequality by a negative number, we have to change the direction of the   $\gt, \: \lt, \: \ge, \: or \: \le$   sign.

The best way to learn about solving inequalities is by example. For each of these, we'll illustrate the solution using a number line. You should learn how to graph the solution sets of inequalities on a number line, too. Note that the solutions to inequalities are sets of solutions, and not just a single number. Here we go ...

### Example 1

Solve:   $3x \gt 8.$

Solution: Just as though this was the equality $3x = 8,$ we begin (and end, actually) by dividing both sides by 3:

\begin{align} 3x &\gt 8 \\[5pt] \frac{3x}{3} &\gt \frac{8}{3} \\[5pt] x &\gt \frac{8}{3} \end{align}

We can sketch a graph of this solution like this: #### Notation

$\frac{8}{3}$ is circled on the graph because it is not included in the set of solutions. The arrow to the right indicates that any number greater than 8/3 will solve the inequality.

#### Set notation

Another way to write the set of solutions is $(\frac{8}{3}, \; \infty),$ which means "The set of all x, from (but not including) 8/3 to infinity." When curved brackets (parentheses) are used, the endpoint is not included in the set. If 8/3 were to be included, we'd write $[\frac{8}{3}, \; \infty].$

### Example 2

Solve:   $2x - 3 \le 1$

Solution: Just as though this was the equality $2x - 3 = 1,$ we begin by adding 3 to both sides, then we divide both sides by 2:

\begin{align} 2x - 3 &\le 1 \\[5pt] 2x &\le 4 \\[5pt] x &\le 2 \end{align}

We can sketch a graph of this solution like this: #### Notation

$2$ is a filled circle on the graph because it is included in the set of solutions. The arrow to the left indicates that any number less than 2 will also solve the inequality.

#### Set notation

Another way to write the set of solutions is $(-\infty, 2],$ which means "The set of all x, from negative infinity up to and including 2." We use a square bracket around the 2 to indicate that 2 is included in the solution set. It is customary to use a round bracket on ∞ because it's not really a number.

### Example 3

Solve:   $-5x + 3 \gt 8$

Solution: Just as though this was the equality $-5x + 3 = 8,$ we begin by subtracting 3 from both sides, then we divide both sides by -5. But remember, when we divide by a negative number, we'll have to flip the direction of the > sign.

\begin{align} -5x + 3 &\gt 8 \\[5pt] -5x &\gt 5 \\[5pt] x &\lt -1 \: \: \color{#E90F89}{\leftarrow \text{div. by -5}} \end{align}

We can sketch a graph of this solution like this: #### Notation

$-1$ is an open circle on the graph because it is not included in the set of solutions. The arrow to the left indicates that any number less than -2 will solve the inequality.

#### Set notation

Another way to write the set of solutions is $(-\infty, -1),$ which means "The set of all x, from negative infinity up to but not including -1."

### Example 4

Solve:   $-4 \lt -5x + 3 \lt 8$

Solution: This is a different kind of inequality, but one that you'll encounter frequently as you study math. It's really just a combination of two single inequalities,

\begin{align} -4 &\lt -5x + 3 \phantom{000} \color{#E90F89}{\text{and}} \\[5pt] -5x + 3 &\lt 8 \end{align}

We can solve those one at a time, but it's convenient to just do both at the same time. The idea is to get x alone in the center of the inequalities. Here are the steps.

First subtract three from every term:

$$-7 \lt -5x \lt 5$$

Now divide all terms by -5, remembering to flip the inequalities:

$$\frac{7}{5} \gt x \gt -1$$

So our solutions are between $x = \frac{7}{5}$ and $x = -1,$ not including the endpoints. A graph of this solution set looks like this: And the set notation equivalent is $x = (-1, \frac{7}{8}),$ where the round brackets indicate that the end points are not part of the solution set.

### Example 5

Solve:   $5(6 + 3x) + 7 \ge 127$

Solution: If this were an equality, $5(6 + 3x) + 7 = 127,$ we'd first move the 7 to the right by subtraction, then distribute the 5 on the left, and continue with standard algebra steps. It's the same for this inequality. Here are the steps:

\begin{align} 5(6 + 3x) + 7 &\ge 127 \\[5pt] 5(6 + 3x) &\ge 120 \\[5pt] 30 + 15x &\ge 120 \\[5pt] 15x \ge 90 \\[5pt] x \ge 6 \end{align}

A graph of this solution set looks like this: And the set notation equivalent is $x = [6, \; \infty),$ where the square bracket indicates that 6 is a part of the solution set.

### Practice problems

Solve the inequality for x, show the set of solutions on a number line and write the solutions in set notation.

 1 $3 \lt 5x + 2x$ Solution \begin{align} 3 &\lt 5x + 2x \\[5pt] 3 &\lt 7x \\[5pt] \frac{3}{7} &\lt x \end{align} $x \in (\frac{3}{7}, \; \infty)$ 2 $9 \ge -2x - 1$ Solution \begin{align} 9 &\ge -2x - 1 \\[5pt] 10 &\ge -2x \\[5pt] -5 &\le x \end{align} $x \in [-5, \; \infty)$ 3 $6 - 4(6x + 7) \ge 122$ Solution \begin{align} 6 - 4(6x + 7) &\ge 122 \\[5pt] -4(6x + 7) &\ge 116 \\[5pt] 6x + 7 &\le -29 \\[5pt] 6x &\le -36 \\[5pt] x &\le -6 \end{align} $x \in (-\infty, \; -6]$ 4 $-138 \ge -6(6x - 7)$ Solution \begin{align} -138 &\ge -6(6x - 7) \\[5pt] 23 &\le 6x - 7 \\[5pt] 30 &\le 6x \\[5pt] 5 &\le x \end{align} $x \in [5, \; \infty)$ 5 $-x \lt 7 (x - 2) - x$ Solution \begin{align} -x &\lt 7 (x - 2) - x \\[5pt] 0 &\lt 7(x - 2) \\[5pt] 0 &\lt x - 2 \\[5pt] 2 &\lt x \end{align} $x \in (2, \; \infty)$ 6 $3(x - 3) - 5x \gt -3x - 6$ Solution \begin{align} 3(x - 3) - 5x &\gt -3x - 6 \\[5pt] 3x - 9 - 5x &\gt -3x - 6 \\[5pt] -2x - 9 &\gt -3x - 6 \\[5pt] x &\gt 3 \end{align} $x \in (3, \; \infty)$ 7 $28 - x \ge 7(x - 4)$ Solution \begin{align} 28 - x &\ge 7(x - 4) \\[5pt] 28 - x &\ge 7x - 28 \\[5pt] -8x &\ge -56 \\[5pt] x &\le 7 \end{align} $x \in (-\infty, \; 7]$ 8 $-2(5 + 6x) \lt 6(8 - 2x)$ Solution \begin{align} -2(5 + 6x) &\lt 6(8 - 2x) \\[5pt] 10 - 12x &\lt 48 - 12x \\[5pt] 10 &\lt 48 \end{align} Now this is true for any x (for all x). $x \in (-\infty, \; \infty)$ 9 $-1 \lt 9 + x \lt 17$ Solution In this case we want to solve for the x in the middle expression. Do this by subtracting 9 from all terms: $$-10 \lt x \lt 8$$ $x \in (-10, \; 8)$ 10 $-3 \le \frac{x}{2} \lt 0$ Solution Multiply all terms by 2 to isolate the x in the middle: $$-6 \le x \lt 0$$ Here's a number-line graph of the solutions $x \in [-6, \; 0)$

 11 $-50 \lt 7x + 6 \lt -8$ Solution \begin{align} -50 \lt 7&x + 6 \lt -8 \\[5pt] -56 \lt 7&x \lt -14 \\[5pt] -8 \lt &x \lt -2 \end{align} $x \in (-8, \; -2)$ 12 $-36 \le 3x - 6 \lt -15$ Solution \begin{align} -36 \le 3&x - 6 \lt -15 \\[5pt] -30 \le 3&x \lt -9 \\[5pt] -10 \le &x \lt -3 \end{align} $x \in (-10, \; -3)$ 13 $-33 \le -7x - 12 \lt -26$ Solution \begin{align} -33 \le -7&x - 12 \lt -26 \\[5pt] -21 \le -7&x \lt -14 \\[5pt] 3 \ge &x \gt 2 \end{align} $x \in (2, \; 3]$ 14 $36 \le 11 - 5x \le 66$ Solution \begin{align} 36 \le 11 - 5&x \le 66 \\[5pt] 25 \le -5&x \le 55 \\[5pt] -5 \ge &x \ge -11 \end{align} $x \in [-11, \; -5]$ 15 $-10 - 2x \lt 6 \: \: \text{and} \: \: 6x + 12 \lt -6$ Solution For an "and" problem like this, we need to solve two inequalities. They are \begin{align} -10 - 2x &\lt 6 \\[5pt] -2x &\lt 16 \\[5pt] x &\gt -8, \: \: \color{#E90F89}{\text{and}} \\[9pt] 6x + 12 &\lt -6 \\[5pt] 6x &\lt -18 \\[5pt] x &\lt -3 \end{align} Putting both solutions together gives us $-8 \lt x \lt -3.$ The number-line graph looks like this: $x \in (-8, \; -3)$ 16 $2x - 3 \lt 11 \: \: \text{or} \: \: -8x - 10 \lt -82$ Solution For an "or" problem like this, we need to solve two inequalities. They are \begin{align} 2x - 3 &\lt 11 \\[5pt] 2x &\lt 14 \\[5pt] x &\lt 7, \: \: \color{#E90F89}{\text{and}} \\[9pt] -8x - 10 \lt -82 \\[5pt] -8x &\lt -72 \\[5pt] x &\gt 9 \end{align} Putting both solutions together gives us $x \lt 7$ or $x \gt 9.$ The number-line graph looks like this: In this case the solution is the union of two sets: $x \in (-\infty, \; 7) \cup (9, \; \infty).$ 17 $4x + 18 \le -12x - 14 \le 14 - 5x$ Solution Because there are variables in each expression in the inequality, the approach here is to solve the first and last inequalities separately: \begin{align} 4x + 18 &\le -12x - 14 \\[5pt] 16x &\le -32 \\[5pt] x &\le -2, \: \: \color{#E90F89}{\text{and}} \\[9pt] -12x - 14 &\le 14 - 5x \\[5pt] -7x &\le 28 \\[5pt] x &\ge -4 \end{align} Putting both solutions together gives us $-4 \le x \le -2.$ The number-line graph looks like this: $x \in [-4, \; -2].$ 18 $4x + 8 \gt 11x + 15 \: \: \text{and} \: \: 13 - 14x \le 13 - 3x$ Solution First we solve the two inequalities separately: \begin{align} 4x + 8 &\gt 11x + 15 \\[5pt] -7x &\gt 7 \\[5pt] x &\lt -1, \: \: \color{#E90F89}{\text{and}} \\[9pt] 13 - 14x &\le 13 - 3x \\[5pt] -11x &\le 0 \\[5pt] x &\ge 0 \end{align} Putting both solutions together gives us $x \lt -1$ and $x \ge 0.$ The number-line graph looks like this: $x \in (-\infty, \; -1) \cup [0, \; \infty).$ 19 $8x - 13 \le -2 + 7x \: \: \text{and} \: \: 20 + 4x \lt 2x + 14$ Solution First we solve the two inequalities separately: \begin{align} 8x - 13 &\le -2 + 7x \\[5pt] x &\le 11, \: \: \color{#E90F89}{\text{and}} \\[9pt] 20 + 4x &\lt 2x + 14 \\[5pt] 2x &\lt -6 \\[5pt] x &\lt -3 \end{align} Now both of these solutions have x less than a number. The only set of solutions that solves the complete inequality is the overlap between the two: The overlap is the set $x \in (-\infty, \; -3).$ 20 $5x + 10 \le -4x - 17 \lt 9 - 2x$ Solution First we solve the left and right inequalities separately: \begin{align} 5x + 10 &\le -4x - 17 \\[5pt] 9x &\le -27 \\[5pt] x &\le -3, \: \: \color{#E90F89}{\text{and}} \\[9pt] -4x - 17 &\lt 9 - 2x \\[5pt] -6x &\lt 26 \\[5pt] x &\lt -\frac{13}{3} \end{align} Now both of these solutions are plotted on the number line below. The only numbers that solve both inequalities are in the overlapping area. The overlap is the set $x \in (-\frac{13}{3}, \; -3].$  xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012-2019, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.