Here is an old party trick. Think of a number. Double it. Add ten. Halve the result. Subtract the number you first thought of. Whatever number you chose, you are now holding 5.

Try it with 7: \(7 \to 14 \to 24 \to 12 \to 5\). With 100: \(100 \to 200 \to 210 \to 105 \to 5\). With \(-3\): \(-3 \to -6 \to 4 \to 2 \to 5\).

Three examples are not an explanation. How do we know it works for every number, including ones nobody will ever try? Arithmetic can only check one number at a time. We need a way to compute with "whatever number you thought of" as a single object.

Give the unknown number a name. Call it \(n\). Then do exactly the same operations, writing down what you have at each stage instead of a specific value:

The letter has done something no list of examples could do: it has carried out the arithmetic for all numbers at once. The halving step is where the magic happens: half of \(2n + 10\) is \(n + 5\), because halving each part of a sum halves the whole. Subtracting \(n\) then removes all trace of the original choice.

That is the entire spirit of algebra. A letter is a placeholder that lets you write down a calculation without committing to the input, so you can reason about the calculation itself.

Arithmetic facts like \(3 \times (4 + 5) = 3 \times 4 + 3 \times 5\) are true for specific numbers. But you will notice that the shape of such facts never depends on the particular numbers. Algebra promotes those shapes to first-class statements: \(a(b + c) = ab + ac\) for all numbers \(a, b, c\). Once a shape is established, you can use it anywhere, on numbers or on other expressions, without checking it again. The handful of rules that make this work (commutativity, associativity, distributivity, and the behaviour of 0 and 1) were reviewed in Chapter 1. Algebra is just those rules applied to symbols.

Look at the equation of a straight line, which you will meet properly in Chapter 5:

Four letters. Are they all "variables"? If you ask a mathematician, she will say \(x\) and \(y\) are variables, while \(m\) and \(b\) are…​ something else. Yet all four are letters standing for numbers. What is the difference, and why does it matter?

The difference is not in the letters but in the role each letter plays in a particular discussion. A useful mental model: think of a family of programs.

Change \(m\) or \(b\) and you get a different line. Change \(x\) and you move along the same line.

Roles can change. In the quadratic formula you will treat \(a, b, c\) as fixed and solve for \(x\). But if you later ask "for which values of \(c\) does this equation have a solution?", \(c\) has become the variable of interest. Mathematicians switch roles constantly and rarely announce it. The context, and phrases like "for fixed \(a\)" or "solve for \(t\)", tell you who is playing which part.

To manipulate something you need names for its parts. A programmer who cannot say "argument", "return value", or "loop body" cannot discuss code. Algebra has the same need, and the vocabulary is small. Here is a typical expression, labelled:

Algebra has a few writing conventions that are so ingrained nobody mentions them. After a gap they can trip you up.

Compare these four lines:

They look similar, but they are four different kinds of statement. Line (a) asserts nothing. Line (b) is true for exactly one value of \(x\). Line (c) is true for every value of \(x\). Line (d) describes a relationship between three quantities. Confusing these kinds is the root of a surprising number of algebraic blunders, so it pays to have precise names.

The same idea in table form:

The reliable test is to simplify both sides and compare.

You will see these three outcomes again, under the names "one solution", "no solution", and "infinitely many solutions", for linear equations in Chapter 5 and for systems of equations in Chapter 6.

A formula is only useful if you can feed values into it reliably. The rules are simple, yet substitution is where most sign errors in all of mathematics are born. Consider evaluating \(3x^2 - 2xy + y^2\) when \(x = -2\) and \(y = 3\). A hurried calculation gives \(-12 - 12 + 9 = -15\). The correct answer is \(33\). Where did the hurried version go wrong?

Substitution means "replace each occurrence of the letter by the value". The danger is that a value is a whole object, while the expression around the letter may bind more tightly than you expect. \(x^2\) with \(x = -2\) means \((-2)^2 = 4\), not \(-2^2 = -4\). The cure is mechanical: always substitute inside brackets. Brackets cost nothing and they protect the value.

Evaluate \(3x^2 - 2xy + y^2\) at \(x = -2\), \(y = 3\).

The hurried version made two errors: it squared \(-2\) to get \(-4\) (no brackets), and it lost the double negative in \(-2xy\). Writing every bracket prevents both.

You can substitute an expression, not just a number. This is how one formula is fed into another. If \(y = x^2 + 2x\) and \(x = t - 1\), then

The brackets around \(t - 1\) are not optional: without them, \(x^2\) would become \(t - 1^2\), which is a different expression entirely. Substitution of expressions is the algebraic face of function composition, which you will meet in Chapter 11.

Try it: −x² in Excel and Python.

The order of operations can feel like an arbitrary rule to memorise. It is better understood as a description of a tree. Every expression has a hidden structure in which each operation combines exactly two sub-expressions. Here is \(3(x + 2) - 4y\):

To evaluate, start at the leaves (numbers and variables), replace each variable by its value, and work upward: each internal node applies its operation to the values of its children. The root’s value is the value of the expression. "Brackets first, then exponents, then multiplication, then addition" is simply the convention for how a flat line of symbols encodes such a tree. Brackets exist to override the default tree shape. Try it: Draw the tree of 3(x+2)−4y.

Substitution is now easy to picture: replace a leaf labelled \(x\) by the value, or by an entire subtree.

Quick: what is \(7 \times 38\)? Most people who can do this in their heads do it as \(7 \times 40 - 7 \times 2 = 280 - 14 = 266\). They have used, perhaps without naming it, the single most important law in algebra. Every time you "multiply out brackets", you are using it. Every time you factorise, you are using it backwards. Let us make it explicit and see why it is true.

Think of multiplication as the area of a rectangle. A rectangle of height 3 and width \(x + 2\) can be cut into two rectangles: one \(3\) by \(x\), one \(3\) by \(2\). The total area is unchanged by cutting, so \(3(x + 2) = 3x + 6\).

The same picture handles two brackets. A rectangle of width \(x + 2\) and height \(x + 3\) cuts into four pieces, one for each pairing of a width-part with a height-part:

Adding the four areas gives \(x^2 + 2x + 3x + 6 = x^2 + 5x + 6\). The picture explains the rule you may remember as "multiply everything in the first bracket by everything in the second".

The distributive law is one of the basic properties of numbers:

It holds for every number, positive or negative, whole or fractional; the area picture is a convincing illustration for positive lengths, and the rules of signed arithmetic in Chapter 1 extend it to all numbers. Because \(b - c = b + (-c)\), it also covers subtraction: \(a(b - c) = ab - ac\). And because it holds for any \(a\), we may let \(a\) itself be a bracketed expression. That gives two-bracket expansion with no new rule at all:

Example 1. A negative multiplier. Expand \(-3(2x - 5)\).

Example 2. Two binomials with signs. Expand \((2x - 1)(x + 4)\). There are \(2 \times 2 = 4\) products:

Check by substitution. With \(x = 1\): the original is \((1)(5) = 5\); the expansion is \(2 + 7 - 4 = 5\). Agreement at one point does not prove the expansion right (see the Watch out in Section 4.4), but disagreement would prove it wrong, so this is a quick, worthwhile sanity check.

Example 3. A binomial times a trinomial. Expand \((x + 1)(x^2 - x + 1)\). There are \(2 \times 3 = 6\) products:

That cancellation is not luck: it is the "sum of cubes" factorisation \(x^3 + 1 = (x + 1)(x^2 - x + 1)\), which you will meet again in Chapter 7.

Three expansions appear so often that it is worth recognising them instantly, in both directions.

Each one is just the four-product rule. For the first: \((a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2\). For the third, the cross terms \(-ab\) and \(+ab\) cancel. In the area model, \((a + b)^2\) is a square of side \(a + b\) cut into an \(a^2\) square, a \(b^2\) square, and two \(ab\) rectangles.

Rule of thumb: spot a special product if you can; use "each by each" (or FOIL) for two binomials; use a grid for anything larger.

Expanding usually leaves a mess: \(4(2x - y) - 3(x - 2y) + 5\) becomes a string of six terms. We want the simplest equivalent form. What does "simplest" mean, and which combinations are legal?

\(3x + 5x = 8x\) because three apples plus five apples is eight apples. But the real justification is the distributive law read backwards:

That explains why only like terms combine. \(3x + 5x^2\) has no common variable part to pull out, so there is nothing to add: it is already as simple as it gets. Apples and apple-squares do not merge.

Simplify \(4(2x - y) - 3(x - 2y) + 5\).

Check at \(x = 1, y = 1\): original \(4(1) - 3(-1) + 5 = 4 + 3 + 5 = 12\); result \(5 + 2 + 5 = 12\). Consistent.

Expanding turns products into sums. Often we need to go the other way. Suppose a shop sells items at \(12\) dollars, \(18\) dollars, and \(30\) dollars, and you buy \(n\) of each: the total \(12n + 18n + 30n\) is obviously \(60n\). Less obviously, \(6x^2 + 9x\) can be rewritten as \(3x(2x + 3)\). Why would you want to? Because a product tells you things a sum hides. A product is zero exactly when one of its factors is zero, so \(3x(2x + 3) = 0\) immediately gives \(x = 0\) or \(x = -\tfrac{3}{2}\); the sum \(6x^2 + 9x = 0\) tells you nothing at a glance. Products also cancel in fractions, and sums do not.

Factoring is the distributive law run backwards: \(ab + ac = a(b + c)\). Look for something every term shares, pull it out to the front, and write what is left of each term inside a bracket. In the area picture, you are taking several rectangles of the same height and pushing them together into one long rectangle.

Factor \(12x^3y - 18x^2y^2 + 6x^2y\).

Check: \(6x^2y \cdot 2x = 12x^3y\), \(6x^2y \cdot (-3y) = -18x^2y^2\), \(6x^2y \cdot 1 = 6x^2y\). Correct, and three terms in, three terms out.

Factoring out a negative. \(-4x + 8\) can be written \(4(-x + 2)\) or \(-4(x - 2)\). Both are correct. The second is usually preferred because it makes the bracket’s leading coefficient positive, which makes common brackets easier to spot, for instance in \(x(x - 2) - 4(x - 2) = (x - 4)(x - 2)\). That last move, pulling out a common bracket, is the seed of factoring by grouping in Chapter 7.

Taking out a common factor is the first and simplest kind of factoring. Recognising \(x^2 - 9\) as \((x - 3)(x + 3)\) (difference of two squares) and \(x^2 + 5x + 6\) as \((x + 2)(x + 3)\) are harder skills developed in Chapter 7 and Chapter 8. The habit to form now: when faced with a sum, always ask first, "is there a common factor?"

So far we have transformed expressions into equivalent expressions. With equations we want something different: to find the values that make the equation true. The strategy is to transform the equation step by step into something so simple that the answer can be read off, such as \(x = 4\). But which transformations are allowed? If you change an equation, how do you know the new one has the same solutions?

Picture an old-fashioned pan balance. An equation says the two pans weigh the same. Whatever you do to one pan, you must do to the other, and the balance stays level. Below, each blue box is an unknown weight \(x\) and each orange disc weighs 1 unit.

Remove three discs from each side: the pans still balance, so \(2x = 8\). Halve the contents of each pan: still level, so \(x = 4\). Check: \(2(4) + 3 = 11\). Correct.

This is al-Khwarizmi’s "balancing", and it is the whole idea of equation solving. Chapter 5 develops it into a complete algorithm.

Which operations produce an equivalent equation? The key requirement is that the step can be undone. If \(A = B\) then certainly \(A + 5 = B + 5\); and conversely, if \(A + 5 = B + 5\) we can subtract \(5\) to recover \(A = B\). Because we can go both ways, no solutions are gained or lost. The safe operations are:

The formula \(F = \tfrac{9}{5}C + 32\) converts a Celsius temperature to Fahrenheit. A thermostat in a US house reads \(68^\circ\text{F}\); what is that in Celsius? You could substitute \(F = 68\) and solve the resulting equation. But if you need to convert a thousand readings, or write a function f_to_c, you want a formula with \(C\) as its subject. Rearranging is equation solving where the "answer" is an expression rather than a number.

Read the formula as a recipe that builds \(F\) from \(C\): start with \(C\), multiply by \(\tfrac{9}{5}\), then add \(32\). To get back from \(F\) to \(C\), undo the steps in reverse order, each with its inverse operation: subtract \(32\), then divide by \(\tfrac{9}{5}\) (that is, multiply by \(\tfrac{5}{9}\)).

It is the same logic as getting undressed: you put on socks and then shoes, so you take off shoes and then socks. Inverses in reverse order.

Example 1. Celsius from Fahrenheit. Make \(C\) the subject of \(F = \tfrac{9}{5}C + 32\).

So \(68^\circ\text{F}\) is \(\tfrac{5}{9}(36) = 20^\circ\text{C}\). Check: \(\tfrac{9}{5}(20) + 32 = 36 + 32 = 68\). Correct.

Example 2. A physics formula. The velocity after constant acceleration is \(v = u + at\) (initial velocity \(u\), acceleration \(a\), time \(t\)). Make \(a\) the subject.

That is the familiar "change in velocity over time". The condition \(t \neq 0\) is not pedantry: at \(t = 0\) no time has passed and the acceleration cannot be deduced.

Example 3. A trapezoid. The area of a trapezoid with parallel sides \(a, b\) and height \(h\) is \(A = \tfrac{1}{2}h(a + b)\). Make \(b\) the subject.

Here we did not expand \(h(a + b)\): dividing by \(h\) was cleaner. Expanding first (\(2A = ha + hb\), then \(hb = 2A - ha\), then \(b = \tfrac{2A - ha}{h}\)) gives the same answer in a different form. Both are correct; choose whichever path keeps the algebra short.

Example 4. The subject appears twice. Make \(x\) the subject of \(y = \dfrac{x + 1}{x - 2}\).

This is the important case. The variable appears in two places, so no single "undo" chain will isolate it. The procedure handles it: clear the fraction, gather the \(x\) terms, factor.

Check with \(x = 5\): \(y = \tfrac{6}{3} = 2\), and the new formula gives \(x = \tfrac{1 + 4}{1} = 5\). Correct. The condition \(y \neq 1\) has meaning: no \(x\) makes \(\tfrac{x + 1}{x - 2}\) equal to 1 (that would require \(x + 1 = x - 2\), i.e. \(1 = -2\)). In the language of Chapter 11, you have just computed an inverse function and found a value outside its domain.

Example 5. A root. A pendulum’s period is \(T = 2\pi\sqrt{L/g}\). Make \(L\) the subject.

With \(g \approx 9.81\ \text{m/s}^2\) and a desired period of \(2\) seconds, \(L \approx 9.81 \times 4 / 39.48 \approx 0.994\) m: a "seconds pendulum" is almost exactly a metre long. (That near-coincidence is not an accident; it was once proposed as the definition of the metre.)

Real problems arrive in words, not symbols. "A taxi charges a fixed fee of $3 plus $1.80 per mile. How far can you go for $30?" The mathematics here is trivial once written down; the hard part is the writing down. Translation is a skill in its own right, and it has a few traps that catch even fluent readers.

Example 1. The taxi. Let \(m\) be the distance in miles. The cost in dollars is \(3 + 1.80m\). A budget of $30 means

You can travel 15 miles. Check in words: $3 plus 15 miles at $1.80 is \(3 + 27 = 30\) dollars. Correct. The expression \(3 + 1.80m\) is a linear cost model, a fixed cost plus a rate; Chapter 5 builds a whole toolkit around models of this shape.

Example 2. Consecutive integers. Three consecutive integers add to 87. Find them.

Let the smallest be \(n\); the others are \(n + 1\) and \(n + 2\).

The integers are \(28, 29, 30\). Check: \(28 + 29 + 30 = 87\).

Example 3. "Less than" reverses. "Five less than twice a number is 17." The phrase "five less than \(X\)" means \(X - 5\), so the equation is \(2n - 5 = 17\), giving \(2n = 22\), \(n = 11\). Writing \(5 - 2n = 17\) is the classic mistake: English puts the 5 first, algebra puts it last.

A programmer reading algebra brings strong habits from code, and some of them mislead. The line

is perfectly ordinary in Python, C, or Java. As a mathematical equation it is a contradiction: subtract \(x\) from both sides and you get \(0 = 1\). Same symbols, entirely different meanings. Where exactly do the two notions of "variable" and "equals" part company?

Within a single mathematical argument, a variable never changes value. When we wrote \(2x + 3 = 11\), then \(2x = 8\), then \(x = 4\), we did not change \(x\): we produced a sequence of equivalent statements about the same unknown. In code, by contrast, the same name can hold different values at different moments, which is why reasoning about imperative programs requires talking about time ("after line 3, x is 6").