For example, consider the equation
2*x* − 4 = 3*x* − 9.
If you substitute *x* := 4 (say),
then you get this result:

- 2
*x*− 4 = 3*x*− 9 — original equation; - 2(4) − 4 = 3(4) − 9 —
substitute
*x*:= 4; - 8 − 4 = 12 − 9 — work out 2 · 4 = 8 and 3 · 4 = 12;
- 4 = 3 — work out 8 − 4 = 4 and 12 − 9 = 3;
- False — since in fact 4 > 3.

- 2
*x*− 4 = 3*x*− 9 — original equation; - 2(5) − 4 = 3(5) − 9 —
substitute
*x*:= 5; - 10 − 4 = 15 − 9 — work out 2 · 5 = 10 and 3 · 5 = 15;
- 6 = 6 — work out 10 − 4 = 6 and 15 − 9 = 6;
- True — since in fact 6 = 6.

For another example, consider the equation *t* = 1/*t*.
Let me try *t* := 2, *t* := 1, and *t* := 0.

*t*= 1/*t*— original equation;- 2 = 1/2 — substitute
*t*:= 2; - False — since in fact 2 > 1/2.

*t*= 1/*t*— original equation;- 1 = 1/1 — substitute
*t*:= 1; - 1 = 1 — work out 1/1 = 1;
- True — since in fact 1 = 1.

*t*= 1/*t*— original equation;- 0 = 1/0 — substitute
*t*:= 0; - Undefined — since 1/0 is undefined.

For the most part,
a meaningless equation is just as good or bad a false one,
and in fact some mathematicians
would count this equation as false when *t* := 0.
However, the difference can be important when we get to inequalities.

An equation may have just one solution, many solutions, or none at all.
As it turns out, *x* := 5
is the *only* solution
to 2*x* − 4 = 3*x* − 9.
However, *t* := 1
is *not* the only solution to *t* = 1/*t*.
This is because *t* := −1 is *also* a solution:

*t*= 1/*t*— original equation;- −1 = 1/(−1) —
substitute
*t*:= −1; - −1 = −1 — work out 1/(−1) = −1;
- True — since in fact −1 = −1.

For an example of an equation with *no* solutions,
consider *x* = *x* + 1;
no matter what real number *x* is,
*x* will in fact always be less than *x* + 1,
so nothing could possibly be a solution to this equation.
For an example of an equation with *many* solutions,
consider the equation *x* = |*x*|;
one solution is *x* := 0,
but in fact any positive number could be used instead.
(On the other hand, no negative number gives a solution.)
Or consider the equation *x* = *x*.
No matter what real number you substitute for *x*,
this statement will obviously be true!
Finally, consider the equation 1/*x* = 1/*x*.
Obviously this statement is true whenever its defined,
but *x* := 0 is not a solution,
because the result is undefined in that case.

So in summary, an equation might have no solutions, one real number as a solution, a few solutions, a whole range of solutions, every real number as a solution, or every real number with one or a few exceptions. Pretty much anything is possible if you pick the right equation!

Consider the possible inequalities
relating the expressions 2*x* − 4 and 3*x* − 9.
Let's evaluate these five inequalities,
each at *x* := 4 and each at *x* := 5.
I already evaluated the relevant expressions above,
so I'll just put in a table of the results:

Inequality: | When x := 4: |
Result: | When x := 5: | Result: |
---|---|---|---|---|

2x − 4 < 3x − 9; |
4 < 3; | False; | 6 < 6; | False. |

2x − 4 > 3x − 9; |
4 > 3; | True; | 6 > 6; | False. |

2x − 4 ≤ 3x − 9; |
4 ≤ 3; | False; | 6 ≤ 6; | True. |

2x − 4 ≥ 3x − 9; |
4 ≥ 3; | True; | 6 ≥ 6; | True. |

2x − 4 ≠ 3x − 9; |
4 ≠ 3; | True; | 6 ≠ 6; | False. |

In a couple of days,
I'll discuss how you would come up with that;
that is, how to be *sure* that *x* := 5 is the only solution.
But for now, consider how to solve the equation *t* = 1/*t*.
If you trust what I said before,
then you know that the only solutions to this equation
are *t* := 1 and *t* := −1.
But you can't replace this equation with the equation *t* = 1,
nor can you replace it with *t* = −1,
because each of these equations is missing one of the solutions.
Instead, you must replace it with this *compound* statement:
*t* = 1 or *t* = −1.

A **compound statement**
consists of simpler statements (like equations or inequalities)
joined by either the word ‘and’ or the word ‘or’.
(If you study Logic,
you'll see that there are other words that might be used here,
like ‘if’ or ‘but not’.
However, we won't need such connectives 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**,
which are inequalities joined with ‘and’,
like *x* > 2 and *x* ≤ 3 above.
This example can be summarised simply as 2 < *x* ≤ 3.
To see this,
first notice that *x* > 2
means the same thing as 2 < *x*,
then use the convention that a bunch of statments run together
always have implicit ‘and’s between them.

Here are some more examples:

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

2 < x ≤ 3; |
2 < x and x ≤ 3; |

−1 < y < 6; |
−1 < y and y < 6; |

5 > a ≥ 3; |
5 > a and a ≥ 3; |

2 < 5x − 9 ≤ 3; |
2 < 5x − 9 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
always use ‘or’ in an *inclusive* sense,
so that if both statements are true, then the compound statement is true.
For example, consider the compond statement
*x* = 2 or *x* < 3.
This is really just equivalent to the simple statement *x* < 3,
because both parts are true when *x* := 2,
so the first part (*x* = 2) is superflouous.

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