So that's how I write the solutions of equations with multiple solutions.t= 1 ort= −1.

In general,
a **compound statement**
consists of simpler statements (such as equations or inequalities)
joined by either AND or OR.
(If you study Logic,
you'll see that there are other connectives that might be used here,
such as IF or BUT NOT.
However, we won't need these in this course.)
For example:

*t*= 1 or*t*= −1;*x*= 2 or*x*= 3;*x*= 2 or*y*= 3;*x*< 2 or*x*≥ 3;*x*> 2 and*x*≤ 3;*p*= 1 or*p*= 5 or*p*= 7;*x*> 2 and*x*≤ 3, or*x*> 4.

Particularly common are **compound inequalities**,
each of which consists of two inequalities joined with AND
and sharing one expression.
An example is

above. This example can be summarised simply asx> 2 andx≤ 3

2 <that is its form as a compound inequality. (Notice thatx≤ 3;

Here are some more examples:

Compound inequality: | Expanded meaning: |
---|---|

2 < x ≤ 3; |
x > 2 and x ≤ 3; |

−1 < y < 6; |
y > −1 and y < 6; |

5 > a ≥ 3; |
a < 5 and a ≥ 3; |

2 < 5x − 9 ≤ 3; |
5x − 9 > 2 and
5x − 9 ≤ 3. |

The words AND and OR
are pretty fundamental English words,
but for purposes of Mathematics, I should tell you exactly what they mean.
A compound statement with AND
is true only if *both* statements are true,
and false if *either* statement (or both!) is false.
Conversely, a compound statement with OR
is true if *either* statement (or both!) is true,
and false only if *both* statements are false.

In particular, mathematicians use OR in an *inclusive* sense,
so that if both statements are true, then the compound statement is true.
For example, consider the compound statement
*x* ≤ 2 or *x* > 1;
this statement is simply True.
Sometimes this is because the first part (*x* ≤ 2) is true,
sometimes it's because the last part (*x* > 1) is true,
and sometimes it's because they are both true
(which happens when 1 < *x* ≤ 2).
But one way or another, no matter what value *x* takes,
at least one of the two parts is true, so the compound statement is true.

- If there is a limited range of values,
as we have with a compound inequality,
give the first value and the last value, separated by a comma;
put round parentheses or square brackets around this pair,
depending on whether these values are or are not included.
For example, the solution set for
*x*of −3 <*x*≤ 2 is (−3, 2]; the round parenthesis around −3 indicates that −3 is not included, while the square bracket around 2 indicates that 2 is included. - If there is an unlimited range of values in the negative direction,
then use the symbol −∞
(pronounced ‘minus infinity’)
as the first value;
always use a round parenthesis there
(since −∞ itself is not a real number,
so it can't possibly be a solution).
For example, the solution set for
*y*of*y*< 3 is (−∞, 3), while the solution set for*y*of*y*≤ 3 is (−∞, 3]. - If there is an unlimited range of values in the positive direction,
then use the symbol
+∞ (pronounced ‘positive infinity’)
or just ∞ (pronounced ‘infinity’)
as the last value;
again always use a round parenthesis there.
For example, the solution set for
*y*of*y*> 3 is (3, ∞), while the solution set for*y*of*y*≥ 3 is [3, ∞). - If there are two or more disjoint ranges of values,
then list them all,
separated by the symbol ∪ (pronounced ‘union’);
it's helpful to list these in increasing order.
For example, the solution set for
*x*of*x*≤ −3 or*x*> 2 is (−∞, −3] ∪ (2, ∞).

Compare interval notation to graphs; you'll see that (except for −∞ and ∞) the round parentheses and square brackets match up perfectly; they're used in the same places, facing in the same directions. This is no coincidence, of course; the notation is designed to work this way!

- 5
*y*+ 4 < 2*y*− 2 ― original inequality; - 3
*y*+ 4 < −2 ― subtract 2*y*from both sides; - 3
*y*< −6 ― subtract 4 from both sides; *y*< −2 ― divide both sides by 3.

Here's an example where I have to change the direction of the inequality:

- 2
*c*+ 5 ≥ 3*c*+ 7 ― original inequality; - −
*c*+ 5 ≥ 7 ― subtract 3*c*from both sides; - −
*c*≥ 2 ― subtract 5 from both sides; *c*≤ −2 ― take the opposite of both sides.

To solve compound statements, you can just treat each statement separately. However, if you have a compound inequality, such as

2 <then instead of breaking this up into 2 <x+ 1 ≤ 5,

1 <that's the answer. To check whenx≤ 4;

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