An equation is a statement claiming that two algebraic expressions are equal. This is indicated simply by placing an equals sign ‘=’ between the expressions.

Evaluating equations

Recall that you evaluate an expression by substituting specific numbers for each of the variables; when you work it out, the result is another specific number (assuming that it's defined). You evaluate an equation in much the same way, by substituting specific numbers for each of the variables; however now, when you work it out, you will get a numerical statement that is either true or false (assuming again that it's defined).

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

On the other hand, if you substitute x := 5, then you get this result: So the equation 2x − 4 = 3x − 9 is false when x := 4 but true when x := 5.

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

So the equation is false when t := 2. So the equation is true when t := 1. So the equation is meaningless (its truth value is undefined) when t := 0.

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.


A solution to an equation is an assignment of values for the variables in the equation such that the equation becomes true when evaluated at those values. For example, we've seen above that x := 5 is a solution to 2x − 4 = 3x − 9, while x := 4 is not a solution of that equation. Also, we've seen that t := 1 is a solution to t = 1/t, while t := 2 and t := 0 are not solutions (albeit for different reasons).

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

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!


An inequality is a statement that two numbers are unequal or ordered in a particular way. It's just like an equation, except that it uses one of the inequality symbols ‘<’, ‘>’, ‘≤’, ‘≥’, or ‘≠’ instead of an equals sign.

Consider the possible inequalities relating the expressions 2x − 4 and 3x − 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.
So x := 4 is a solution of the inequalities 2x − 4 > 3x − 9, 2x − 4 ≥ 3x − 9, and 2x − 4 ≠ 3x − 9, while x := 5 is a solution of the inequalities 2x − 4 ≤ 3x − 9 and 2x − 4 ≥ 3x − 9 (and also, as we saw earlier, of the equation 2x − 4 = 3x − 9). All of these inequalites, however, have many solutions, which is what you should expect for an inequality.

Compound statements

Much of algebra involves solving equations (and inequalities). To solve an equation, you replace it with a simpler statement that has exactly the same solutions. For example, to solve 2x −4 = 3x − 9, you would replace it with x = 5, since x := 5 is the unique solution to each of these equations.

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:

The last couple of examples are extra-compound; for example, to form the last statement, you first join ‘x > 2’ to ‘x ≤ 3’ with ‘and’, then join this compound statement to ‘x > 4’ with ‘or’. But statements that complicated are pretty much never going to come up in this course.

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.
In this last example, I put used a complicated algebraic expression, just to show that you can do this; some problems in the future will have something like this, but it's not considered to be solved (and you can always avoid it if you want to). Even the example before it, 5 > a ≥ 3, is not really solved in the way that we usually like; it's preferable to write it as 3 ≤ a < 5.

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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