Making Sense of Algebra

This is a first pass introduction to Elementary Algebra, also known as High School Algebra, College Algebra, or just Algebra. The target audience is anyone who has never taken the subject and wants to know what it's about. It also acts as a refresher of the main ideas for adults who have taken the topic back in high school and have forgotten it since. The emphasis is on a deep and detailed understanding of what the topic is and what it represents, not just blindly following rules or mindlessly drilling problems.

Who cares?

Arithmetic is the usage of the basic operators (addition, subtraction, multiplication, and division) to navigate and manipulate the space of numbers. In the real world, we might lack access to the specific numbers, as they could be constantly changing or hard to measure accurately. Consider this scenario:

Example 1

You're saving up to buy a phone. It costs €380, and you've already saved up €80. You can save up an extra €15 every week. In how many weeks can you afford the phone?

You can simply wait and measure the amount of weeks it takes you, but is it possible to predict the outcome mathematically before it even happens? Elementary Algebra is the method that humans developed to find answers to these kinds of problems. Let's build up each idea to see if we can solve it ourselves.

We can simplify this problem if we represent it in a way that's more easily transformed and manipulated. We start by translating it from day to day language, to the language of algebra.
First we learn the basics of how that language works.

Statements, Expressions, and Equations

When we encounter a sentence like "A triangle has three sides.", what can we say about it?
The sentence is talking about a mathematical object, and we know for certain that it's true. It follows directly from the definition of triangles.

A statement is a sentence or a mathematical expression that is either definitely true or definitely false, correct or incorrect; thus statements are pieces of information that we might apply logic to in order to produce other pieces of information (which are also statements).
That's why "A triangle has three sides." is an example of a statement, another example is "The factors of 15 are 5 and 3."
The sentence "The door on the left." isn't a statement,
and neither is
"The number 11."
as these are just references to objects; they aren't sentences concerning a fact.
If you can meaningfully ask "Is this true?" then it's a statement.

Mathematical statements are composed of expressions, which are arrangements of symbols that refer to a specific object.
\(16 + 5\) is an expression, and it evaluates to \(21\) which is also an expression.
On their own, these aren't statements, because they can't be assigned true or false values. What would it even mean for a number to be true or false?
This is analogous to day to day language, where a noun phrase refers to an object, and a whole sentence refers to a fact.
The following are examples of expressions:
\(\sqrt{2} + 9\)
\(3 \times 15 - 11\)
"The result of multiplying \(3\) and \(5\)."
We'll see more examples of expressions and how they help us later on.

A special type of sentence is an equation, here's what one can look like:
\(50\) is equal to \(25 + 25\). That's a true statement.
Notice that it's composed of two expressions, \(50\) and \(25 + 25\) with 'is equal to' connecting them.
We abbreviate "is equal to" with the symbol \(=\), because it's cumbersome to write every time.
\(17 = 16 + 5\) is a false statement.
If we negate that statement by adding a 'not' in front of the 'equal to', the result is called an inequality and is no longer considered an equation:
\(17\) is not equal to \(16 + 5\)
or
\(17 \ne 16 + 5\).
The truth value gets flipped, and it becomes a true statement.

To make this more relevant to our problem, we now bring up algebraic expressions.

Variables, Algebraic Terms, and Polynomials

We need to find the number of additional weeks it'll take to afford the phone. That number could be anything, so it would be very handy to have a way of referring to that unknown quantity, so that we can build up mathematical expressions with it.
We could use the word "weeks", but mathematicians like to be terse, so we abbreviate it to '\(w\)'. '\(w\)' is now what we call a 'variable', it's a symbol referring to an unknown quantity, or a quantity with an unspecified/changing value.

This is what's called abstraction, and it's one of the most critical ideas of algebra. We abstract away specific numbers with letters, and the letters become representations of the general idea of a number (or other object), without referring to a specific one.
In much the same way when we use the symbol \(3\), we refer to a general notion of "threeness" and not a specific triplet of things like sheep, or rocks. This lets us talk about the properties of the number \(3\), and now we can talk about the properties of the variable '\(w\)' which is an abstraction over all possible numbers.
The letters \(x\), \(y\), and \(z\) are most frequently used to refer to generic variables.

We can now build up an expression using our variable '\(w\)', to represent the facts of the problem mathematically in a way that will help us solve it.

You can save up an extra €15 every week.

We're given the fact that we save up an additional \(€15\) every week. That means we can multiply the number of weeks we've been saving up by \(15\) to get the number of euros we've accumulated.
\(15w\) is an algebraic expression that represents that idea.
It's algebraic because it involves a variable combined with the arithmetic operations. The \(15\) there is called a coefficient which means it's a specific value that is multiplied by a variable. When we multiply coefficients and variables together, it's conventional to omit the multiplication symbol to save space, but it just means "\(15\) multiplied by \(w\)".

...you've already saved up €80.

This is the second fact we can incorporate into our expression. The fact that we've already saved up a certain amount means we can add that to the previous expression to get the total we've saved up, dependent on the number of weeks.

\(15w + 80\) is now the expression that represents the total amount we've saved.

The 80 in that expression is referred to as a constant term, because it has no variables attached to it.
Each addition or subtraction we add to the expression is referred to as a term, and an expression with multiple terms like
\(2x + 3y + 12z + 4\) is what we call a polynomial (from Greek poly, meaning "many" and Latin nomen meaning "name" or "term").

Substitution

Let's assume for a moment that we're on the second week of saving, that means that \(w = 2\) by definition.
Since a variable is involved in our \(15w + 80\) expression, we can't get a concrete value out of the expression by trying to do the arithmetic.
If we want to get a concrete value out, we need to supply a concrete value into the arithmetic operation. If we assume (or know) that \(w\) has a specific value, we can replace it with that value in the expression we have.
\(15 \times 2 + 80\)
Using the rules of arithmetic, we can then evaluate that expression to be \(110\). This process is called substitution or 'plugging in' a value. We can usually simplify algebraic expressions by plugging in known values, so this is an additional tool in our toolbox.
But the whole problem we're facing is that we don't know what value \(w\) has, that's why we're referring to it using a letter and not a specific number. Is there anything that will let us get closer to this answer?

Constructing and Solving Equations

Constraints

It costs €380...

This is the third important fact that's given to us in the problem, and it's special. It's what's known as a 'constraint', it limits the possible values that our expression \(15w + 80\) can take on.
Specifically it tells us that the value of the expression must be \(380\), as that's the point where we know we can afford the phone's cost. If our expression has a value below that, we can't afford it, and if it's above that then we know we've saved up for a longer time than necessary.
In other words, we can use this constraint to set up the following algebraic equation:
\(15w + 80 = 380\)
Is this a statement?
Not on its own, since the truth value of it depends on whatever the value of \(w\) is. If \(w = 3\) then the equation becomes false, as you can check for yourself by substitution.
This is what we call an open sentence, a sentence whose truth value depends on a variable.

Recall that the truth value of the statement is a given.
The nature of it being true is a premise built into the problem itself, it's a constraint on what we know. We can say that we're trying to find the \(w\) value that makes the equation true.
This means we can reframe the original problem in the following abstract way:

If \(15w + 80 = 380\) is a true statement, what is the value of \(w\)?

The beauty of algebra is that it gives us the tools to answer that kind of question.
When we assert \(15w + 80 = 380\) is true, by logical necessity the value of \(w\) collapses into a single value, that we can find through applying the properties of numbers in a logical chain of reasoning. Doing this is called 'solving' the equation, and we'll see how to do it now.

How to Solve It

This is the real problem solving step, and it's the part that children in schools spend the most time learning how to do. Like most things in maths, there's an art and a finesse to it, and the method only becomes natural and intuitive with a lot of practice. Mathematics isn't a spectator sport. I recommend following through this part yourself with pen and paper. You can also get more practice at Khan Academy

\(15w + 80 = 380\)

Because it's an equation that we're assuming to be true, we know that both sides of the equation are equal in value. A logical consequence of that is that if we change both sides by the same amount, the equation stays true.
If \(15w + 80 = 380\) is true, then adding \(10\) to both sides gives \(15w + 80 + 10 = 380 + 10\), which is also true.
This gives us the ability to isolate the term with the variable in it, by applying arithmetic operations to both sides to remove all the other terms.
Critical: We must remember that any operation we apply must be done to both sides, as to not change the truth value of the equation.
This process is called algebraic manipulation, we're changing the shape of the statement without changing its value or what it represents.

Here's what the series of logical steps looks like:
If \(15w + 80 = 380\) is true, then \(15w + 80 - 80 = 380 - 80\) is also true. \(\rightarrow\) Subtracting \(80\) from both sides, maintaining the equality.
We can then simplify further through arithmetic to get the following:
\(15w = 300\).
If \(15w = 300\) is true, then \(\frac{15w}{15} = \frac{300}{15}\) is also true. \(\rightarrow\) Dividing by \(15\) from both sides, maintaining the equality.
We can then simplify further into:
\(w = 20\).

What we just did is called 'solving an equation in one variable'.
That series of steps is what we call a proof by calculation of the statement "If \( 15w + 80 = 380\), then \(w = 20\)".
It's a proof because we've demonstrated beyond reasonable doubt to anyone reading that the statement is true. The proof itself is a large sentence, so we include a full stop at the end of it. Using these techniques, we've 'solved' for the value of \(w\).

In practice, we don't write it this way every single time, once we have sufficient practice we can start to abbreviate the above into these steps:

\(15w + 80 = 380\)
\(15w = 300\)
\(w = 20.\)

Example 2

Imagine you're going on a trip to the amusement park. The regular admission cost is €40. Today, they have a special offer: you can buy a wristband for an extra €25, which lets you go on unlimited rides for free. Without the wristband, each ride costs €3.50. You want to figure out how many rides you would need to go on for the wristband to be a better deal.

Let's use the techniques we've been learning to translate this problem into an equation.
First, we assign a variable \(r\) to the number of rides and try to build an expression.
If we assume we're not getting the wristband, each ride costs \(€3.50\). The total cost of the trip is going to be the number of rides we go on, multiplied by the cost of each ride, plus the admission cost.
\(3.50r + 40\) is the resulting expression for the total cost without the wristband.

If we get the wristband then the total cost is only the cost of the wristband plus the cost of the admission.
\(25 + 40\) is the resulting expression for the total cost with the wristband.

How do we know which is the better deal? If we only go on one ride, then you can substitute \(r = 1\) and see that the wristband wouldn't be worth the money.
However, there's a point where the number of rides makes the total cost without the wristband equal to the cost with the wristband. If we exceed that ride limit, it's much better off to buy the wristband. The total cost with the wristband represents a constraint on the number of rides.
The phrase 'equal to' gives us a hint that we want to use an equation.
\(3.50r + 40 = 25 + 40 \)

We now have an equation that represents the point where the cost of individual rides equals the cost of the wristband. Let's solve it!

\(3.50r + 40 = 25 + 40\)

What algebraic manipulations do we need to isolate the variable? We can subtract the constant term \(40\) from both sides.
Common Pitfall: We must make sure that any operation we apply must be done to both sides, as to not change the truth value of the equation. If we only do the operation on one side, the equation becomes a false statement.
\(3.50r + 40 - 40 = 25 + 40 - 40\) which simplifies to \(3.50r = 25\).
Because the left hand side only contains the variable multiplied by a coefficient, we can now divide both sides by that coefficient to fully isolate the variable.
\(\frac{3.50r}{3.50} = \frac{25}{3.50}\) which simplifies to \(r = \frac{25}{3.50}\).
That's approximately equal to \(7.1\), and since you can't get a fraction of a ride, we instead take the ceiling. That means we round up the number to the nearest whole number above it, which is \(8\).
After \(8\) rides, the wristband becomes the more cost-effective option. Once again, in practice, the solution would be abbreviated to the following steps:

\(3.50r + 40 = 25 + 40\)
\(3.50r = 25\)
\(r = \frac{25}{3.50}\)
Taking the ceiling,
\(r = 8\).

For the sake of completeness I have to explain that not all equations involving variables are solvable; in some cases, the truth value of the equation is independent of the variable(s) it contains. Consider the following equation:

\(x + 20 = x + 20\)

This equation has an infinite number of solutions, no matter what value you plug in for \(x\) it will be a true statement. These kinds of equations are known as 'identities' and can often use the symbol \(\equiv\) rather than \(=\).
Another type of equation is an equation that has no solution, such as:

\(5x + 7 = 3x + 2x + 4\)

The constraints in this equation are too tight. If we assume that it's true and try and solve it using the method described above we end up with \(7 = 4\) which is false, and thus a logical contradiction.
We assumed something was true, followed through the logical implications, and ended up getting a false nonsensical answer. That means whatever we assumed in the first place must be actually false, so there's no value of \(x\) that will make the equation true.

Try yourself!

You and your friends go to a coffee shop, and you're paying. You all order coffee for €4 each, and you also buy one pastry for €3. Your total bill comes to €19. How many coffees did you buy?

Solution

Tap to reveal.

\(4c + 3 = 19\)
\(4c = 16\)
\(c = 4\).
Four coffees were bought.

The Significance

This problem we've been solving is obviously a toy problem, one which you might have been able to solve in your head without the usage of any of these techniques. The final answer here isn't important; the important part is that we now have a general problem solving method which we can apply to a variety of situations much more complicated than this one. First, abstracting away the unnecessary details of the problem, and using that abstraction to write down a mathematical expression.
Followed by using the constraints of the problem to construct an equation, and finally solving that equation using algebraic techniques.

Algebra is not just a tool for solving classroom exercises, it is a language for describing patterns, relationships, and change. The abstraction and generalization you learn are the same skills used by scientists to model physical phenomena, by engineers to design systems, and by economists to analyze markets. Whenever you see a formula, a graph, or a computer simulation, algebra is working behind the scenes.

Algebra is emphasized so much in maths education because of its ability to turn real-world problems into mathematical ones. It gives us a real foothold to make predictions, optimize solutions, and understand complex systems. Algebra is used in calculating interest rates, predicting population growth, and designing bridges. The techniques you learn here also form the basis for more advanced mathematics, such as calculus, statistics, and linear algebra. I hope I've shown that life can be made easier when we understand how to apply algebraic thinking to day-to-day decisions, especially decisions that involve numbers.

Thanks for reading, and have a great time learning.