What Is Algebra? Letters, Constants, Variables

Consider the following arithmetic calculations

(5 - 3) (5 + 3) = 2 \cdot 8 = 16 = 25 - 9 = 5^{2} - 3^{2}

(8 - 1) (8 + 1) = 7 \cdot 9 = 63 = 64 - 1 = 8^{2} - 1^{2}

(12 - 3) (12 + 3) = 9 \cdot 15 = 135 = 144 - 9 = 1 2^{2} - 3^{2}

Here we see a pattern. To achieve generality we may use letters to
represent unspecified numbers, and write

(a - b) (a + b) = a^{2} - b^{2}

While arithmetic deals with calculations of specified numbers, in
algebra and calculus to express universal facts, we often use letters
to denote numbers in general, not particular numbers.

If the letter represents a specific number that does not change during
a problem, it is called a constant. But if the letter is allowed
to represent different numbers during a single problem, it is called
a variable. Of course, numbers like $2, - 7, π, \sqrt{3}$ are
also constants.

In equations, the unknown is also called the variable. For example,
in the equation $3 x + 2 = 5$ , the letter $x$ is called the variable
although the equation implies that $x$ can take only one value $x = 1$ .
But in an equation like $y = 2 x + 3$ , $x$ and $y$ represent infinitely
many numbers. It is easier to always call the unknown a variable although
it sometimes represents a single value.

If in a single discussion, both constants and variables appear, constants
are usually denoted by the first letters of the alphabet as $a, b, c, . . .$
and variables often by the last letters of the alphabet as $x, y, z$ .
But this is not a hard-and-fast rule, and for example, the following
statements have the same meaning:

(a - b) (a + b) = a^{2} - b^{2} and (x - y) (x + y) = x^{2} - y^{2} .

The fact that a letter is a constant or a variable should be easily
understood from the context of the problem.

Exercises

Exercise 1.

Check the pattern shown at the start of this section for the numbers $9$ and $2$ . That is, work out $(9 - 2) (9 + 2)$ and work out $9^{2} - 2^{2}$ , and see whether the two results agree.

Answer

Both sides equal $77$ .

Solution

We work out each side separately and only compare at the very end.

Left side. First do the subtraction and the addition inside the parentheses, because parentheses are worked out before multiplying:

9 - 2 = 7, 9 + 2 = 11.

So the left side is

(9 - 2) (9 + 2) = 7 \cdot 11 = 77.

Right side. The symbol $9^{2}$ means $9 \cdot 9$ , and $2^{2}$ means $2 \cdot 2$ :

9^{2} = 81, 2^{2} = 4.

So the right side is

9^{2} - 2^{2} = 81 - 4 = 77.

Both sides came out to $77$ , so the two agree. This is one more case of the pattern

(a - b) (a + b) = a^{2} - b^{2}

with $a = 9$ and $b = 2$ .

A warning about a common slip: $9^{2}$ is $81$ , not $18$ . The little $2$ tells you to multiply $9$ by itself, not to double it.

Exercise 2.

In the equation $5 t + 4 = 19$ , which letter is the variable, and which numbers are the constants?

Answer

The variable is $t$ ; the constants are $5$ , $4$ and $19$ .

Solution

This section explains that in an equation, the unknown letter is called the variable, and that a number or a letter standing for a fixed number is called a constant.

In $5 t + 4 = 19$ the only letter is $t$ , and $t$ is the unknown we would want to find. So $t$ is the variable.

Everything else in the equation is a specific number that does not change: $5$ , $4$ and $19$ . Numbers like these are constants, exactly as the section says about $2, - 7, π, \sqrt{3}$ .

One point that confuses many students: here $t$ can only be one number, namely $t = 3$ , because putting $t = 3$ into the left side gives

5 \cdot 3 + 4 = 15 + 4 = 19,

which matches the right side. So $t$ does not really "vary" at all. But we still call it the variable. The section makes this exact point about the equation $3 x + 2 = 5$ , where $x$ can only be $1$ and is still called the variable. The word "variable" is a naming habit, not a promise that the letter takes many values.

Exercise 3.

Use the formula $(a - b) (a + b) = a^{2} - b^{2}$ with $a = 10$ and $b = 6$ to work out the product $4 \cdot 16$ .

Answer

$64$

Solution

The formula has a factor $a - b$ and a factor $a + b$ . To use it we must first see our product $4 \cdot 16$ in that shape.

We are told to take $a = 10$ and $b = 6$ . Then

a - b = 10 - 6 = 4,

a + b = 10 + 6 = 16.

So the product $4 \cdot 16$ really is $(a - b) (a + b)$ for these values of $a$ and $b$ . That is what allows us to use the formula; without this matching step the formula would not apply.

Now replace the left side by the right side of the formula:

4 \cdot 16 = (10 - 6) (10 + 6) = 10^{2} - 6^{2} .

Work out the two squares:

10^{2} = 100, 6^{2} = 36,

10^{2} - 6^{2} = 100 - 36 = 64.

Check. Multiplying directly, $4 \cdot 16 = 64$ . The two answers agree.

Exercise 4.

The section states the same fact in two ways, once with the letters $a, b$ and once with the letters $x, y$ . Write that same fact a third time using the letters $m$ and $n$ . Does the new version say anything different from the first two?

Answer

$(m - n) (m + n) = m^{2} - n^{2}$ . It says exactly the same thing.

Solution

Replace every $a$ by $m$ and every $b$ by $n$ in

(a - b) (a + b) = a^{2} - b^{2}

to get

(m - n) (m + n) = m^{2} - n^{2} .

The new version says nothing new. The section explains why: the letters are only stand-ins for unspecified numbers. The statement is a claim about every pair of numbers, and which letters we use to point at those two numbers has no effect on the claim.

Here is the test that makes this concrete. Pick any two numbers, say $6$ and $1$ . All three versions tell you the same thing:

(6 - 1) (6 + 1) = 6^{2} - 1^{2},

that is, $5 \cdot 7 = 36 - 1$ , or $35 = 35$ .

Notice also the habit the section describes: $a, b, c$ are usually used for constants and $x, y, z$ for variables. That is only a habit, not a rule, and it does not change the meaning of the formula.

Exercise 5.

Work out $53 \times 47$ without doing long multiplication, by first writing the two numbers in the form $a + b$ and $a - b$ .

Answer

$2491$

Solution

The formula in this section handles a product of the shape $(a - b) (a + b)$ : two numbers that sit the same distance on either side of some middle number. So the first job is to find that middle number.

Both $53$ and $47$ are $3$ away from $50$ :

53 = 50 + 3, 47 = 50 - 3.

So we may take $a = 50$ and $b = 3$ . This matching step is what lets us use the formula.

Now rewrite the product and apply $(a - b) (a + b) = a^{2} - b^{2}$ :

53 \times 47 = (50 + 3) (50 - 3),

(50 + 3) (50 - 3) = 50^{2} - 3^{2} .

Work out the two squares:

50^{2} = 2500, 3^{2} = 9,

2500 - 9 = 2491.

Check. Doing it the long way,

53 \times 47 = 53 \times 40 + 53 \times 7,

53 \times 40 = 2120, 53 \times 7 = 371,

2120 + 371 = 2491.

The answers agree.

Note that the order of the two factors does not matter: $(50 + 3) (50 - 3)$ and $(50 - 3) (50 + 3)$ are the same product, since multiplication can be done in either order.

Exercise 6.

A square lawn measures $34$ metres on each side. A square patio measuring $16$ metres on each side is built in one corner of it. How many square metres of lawn are left? Use the formula of this section rather than working out the two squares.

Answer

$900$ square metres.

Solution

The area of a square is the side multiplied by itself, that is, the side squared.

Area of the whole lawn:

34^{2} square metres .

Area taken up by the patio:

16^{2} square metres .

The lawn that is left is the whole lawn minus the patio:

34^{2} - 16^{2} .

This is exactly the right side of the formula

(a - b) (a + b) = a^{2} - b^{2}

with $a = 34$ and $b = 16$ . The formula can be read from right to left just as well as from left to right, since it says the two sides are equal. Reading it from right to left:

34^{2} - 16^{2} = (34 - 16) (34 + 16) .

Now the arithmetic is easy:

34 - 16 = 18, 34 + 16 = 50,

18 \cdot 50 = 900.

So $900$ square metres of lawn are left.

Check. The long way round: $34^{2} = 1156$ and $16^{2} = 256$ , and $1156 - 256 = 900$ . The answers agree, and the short way avoided both large squares.

Exercise 7.

Two whole numbers $a$ and $b$ satisfy $a + b = 20$ and $(a - b) (a + b) = 40$ , with $a$ larger than $b$ . Find $a$ and $b$ , then work out $a^{2} - b^{2}$ .

Answer

$a = 11$ and $b = 9$ ; and $a^{2} - b^{2} = 40$ .

Solution

This problem runs the formula backwards: we are given the answer and asked for the numbers.

Step 1: find $a - b$ . We are told

(a - b) (a + b) = 40,

and we are also told that $a + b = 20$ . Putting $20$ in place of $a + b$ :

(a - b) \cdot 20 = 40.

So the number $a - b$ , when multiplied by $20$ , gives $40$ . The only such number is $2$ , because $2 \cdot 20 = 40$ . Hence

a - b = 2.

Step 2: find the two numbers. We now need two whole numbers that add up to $20$ and differ by $2$ . Try the pairs adding to $20$ , starting near the middle:

$10$ and $10$ differ by $0$ — too small a difference;
$11$ and $9$ differ by $2$ — this works;
$12$ and $8$ differ by $4$ — too large.

Since $a$ is the larger one, $a = 11$ and $b = 9$ .

Step 3: work out $a^{2} - b^{2}$ . By the formula,

a^{2} - b^{2} = (a - b) (a + b) = 2 \cdot 20 = 40.

Check. Directly, $11^{2} = 121$ and $9^{2} = 81$ , and $121 - 81 = 40$ . This matches, and it also matches the $40$ we were given at the start, as it should.

Exercise 8.

A student writes the following two lines:

(7 - 3) (7 + 3) = 7^{2} - 3^{2} = 49 - 9 = 40,

(7 - 3)^{2} = 7^{2} - 3^{2} = 49 - 9 = 40.

One line is correct and one is wrong. Say which is which, work out the correct value of the wrong line, and explain what went wrong.

Solution

The first line is correct. The second line is wrong.

Why the first line is correct. It is the formula

(a - b) (a + b) = a^{2} - b^{2}

used with $a = 7$ and $b = 3$ . The two factors are $7 - 3$ and $7 + 3$ , which is exactly the shape the formula asks for. And the arithmetic checks out: $7 - 3 = 4$ , $7 + 3 = 10$ , and $4 \cdot 10 = 40$ .

Why the second line is wrong. The symbol $(7 - 3)^{2}$ means $(7 - 3)$ multiplied by itself:

(7 - 3)^{2} = (7 - 3) (7 - 3) .

Here both factors are $7 - 3$ . But the formula needs one factor to be $a - b$ and the other to be $a + b$ — a minus in one and a plus in the other. Since $7 - 3$ is not $7 + 3$ , the formula simply does not apply to this expression, so the student was not allowed to use it.

The correct value:

(7 - 3)^{2} = 4^{2} = 16,

not $40$ .

The lesson. Before using a formula, check that what you have really has the shape the formula describes. Here the student saw a subtraction and two squares and reached for the formula without checking the second factor. This particular mistake — treating $(a - b)^{2}$ as though it equalled $a^{2} - b^{2}$ — is one of the most common errors in all of algebra, so it is worth remembering that

(7 - 3)^{2} = 16 while 7^{2} - 3^{2} = 40.

Exercise 9.

Is the following statement always true, sometimes true, or never true? Justify your answer with examples from this section. "A letter that is called the variable stands for more than one number."

Answer

Sometimes true.

Solution

The answer is sometimes true. To justify this we need one case where the statement holds and one case where it fails.

A case where it holds. In

y = 2 x + 3,

the letter $x$ may be given any number at all, and each choice produces a matching $y$ . For instance $x = 0$ gives $y = 3$ ; $x = 1$ gives $y = 5$ ; $x = 10$ gives $y = 23$ . So here $x$ really does stand for infinitely many numbers, and the statement is true.

A case where it fails. In

3 x + 2 = 5,

the letter $x$ is still called the variable, but as the section points out, the equation forces $x$ to be $1$ and nothing else. Putting $x = 1$ into the left side gives $3 \cdot 1 + 2 = 5$ , which matches the right side; any other value of $x$ would not. So here the variable stands for exactly one number, and the statement is false.

Since we have found a case where it is true and a case where it is false, the statement is sometimes true.

What this shows is that "variable" is a name we give to the unknown letter out of convenience, as the section says, and not a guarantee about how many values that letter can take. Compare this with a constant, such as the $3$ and the $2$ in $3 x + 2 = 5$ , which is fixed for the whole problem by definition.

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