Sometimes we're less interested in solving an equality than an *in*equality. 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

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.

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 ...

Solve: $3x \gt 8.$

**Solution**

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

We can sketch a graph of this solution like this:

$\frac{8}{3}$ 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.

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 $[\frac{8}{3}, \; \infty].$

Solve: $2x - 3 \le 1$

**Solution**

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

We can sketch a graph of this solution like this:

$2$ 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.

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.

Solve: $-5x + 3 \gt 8$

**Solution**

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

We can sketch a graph of this solution like this:

$-1$ 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.

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."

Solve: $-4 \lt -5x + 3 \lt 8$

**Solution**

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

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.

Solve: $5(6 + 3x) + 7 \ge 127$

**Solution**

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

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.

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. |
$-1 \lt 9 + x \lt 17$ ## SolutionIn 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. |
$-3 \le \frac{x}{2} \lt 0$ ## SolutionMultiply 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. |
$-10 - 2x \lt 6 \: \: \text{and} \: \: 6x + 12 \lt -6$ ## SolutionFor 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. |
$2x - 3 \lt 11 \: \: \text{or} \: \: -8x - 10 \lt -82$ ## SolutionFor 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. |
$4x + 18 \le -12x - 14 \le 14 - 5x$ ## SolutionBecause 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. |
$4x + 8 \gt 11x + 15 \: \: \text{and} \: \: 13 - 14x \le 13 - 3x$ ## SolutionFirst 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. |
$8x - 13 \le -2 + 7x \: \: \text{and} \: \: 20 + 4x \lt 2x + 14$ ## SolutionFirst 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. |
$5x + 10 \le -4x - 17 \lt 9 - 2x$ ## SolutionFirst 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 $x \in (-\frac{13}{3}, \; -3].$ |

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