Solving Linear Equations

An equation is a condition, not a statement

Read $3x + 5 = 17$ literally and it looks like a claim: three times something, plus five, equals seventeen. But the something is unknown, so the claim is not yet true or false. It becomes a definite claim only once you choose a value for $x$. Put $x = 4$ into it and you get $17 = 17$, which is true. Put $x = 1$ into it and you get $8 = 17$, which is false. An equation in $x$ is therefore not a single statement at all. It is a condition on $x$, a test that each number where the equation is defined either passes or fails. Every equation in this lesson is defined for all real numbers until fractions with $x$ in a denominator show up later on, so for now “each number where it is defined” simply means “each real number.” Once denominators enter, that phrase will start to matter.

The numbers that pass are the only thing solving is ever after. The solution set of an equation is the set of all numbers, from wherever it is defined, that make it true. We write it with set braces, so the solution set of $3x + 5 = 17$ is $\{4\}$, the set whose only member is $4$. Two symbols will recur. The symbol $\mathbb{R}$ stands for the set of all real numbers, every number on the number line. The symbol $\varnothing$ stands for the empty set, the set with no members at all, which is what we write when nothing passes the test.

Sorting equations by their solution set gives exactly three kinds.

An equation is conditional when it is true for some values in its domain but not others, so its solution set is neither empty nor the whole domain. Every equation you ever solved for a numerical answer was conditional. For instance $3x + 5 = 17$ is conditional, with solution set $\{4\}$.

An equation is an identity when it is true for every value in its domain, so its solution set is the whole domain. For an equation defined everywhere, that means solution set $\mathbb{R}$: for instance $2(x + 1) = 2x + 2$ holds no matter what $x$ is, because the two sides are the same expression written two ways. Later in this lesson you will meet equations whose domain is missing a point, where being an identity means holding at every value the domain does include, not at that missing point.

An equation is a contradiction when it is true for no value in its domain, so its solution set is $\varnothing$. For instance $x = x + 1$ can never hold, since a number cannot equal one more than itself.

This reframes what the word solve means. To solve an equation is to describe its solution set, not merely to “find $x$.” For a conditional equation with a single solution the two sound identical, and writing $x = 4$ is a fine shorthand for “the solution set is $\{4\}$.” But for an identity the honest answer is “every value in its domain” (which so far in this lesson means every real number, until denominators enter and shrink the domain), and for a contradiction it is “no solution, $\varnothing$.” An equation with no single number to report still has an answer. The answer is a description of the set.

Equivalent equations and the one safe move

Solving works by replacing a hard equation with an easier one that has the same solution set, then repeating until the set can be read off at a glance. Two equations are equivalent when they have exactly the same solution set. The equation $x = 4$ is equivalent to $3x + 5 = 17$ because both have solution set $\{4\}$. The whole job of solving is to travel from $3x + 5 = 17$ to $x = 4$ through a chain of equivalent equations.

So the central question of this entire chapter is narrow and precise: which operations on an equation are guaranteed to produce an equivalent one? The answer is a single word, reversible. A step you can undo is guaranteed to produce an equivalent equation, and here is why that one word carries the guarantee.

Two familiar moves are reversible for every real number, so both are always safe. Adding the same quantity to both sides is undone by subtracting it, as long as the quantity is defined for every $x$. Adding a constant like $5$, or a whole expression like $3x$, is fine, because $5$ and $3x$ are defined for every real number and can be subtracted straight back. (Adding something like $\frac{1}{x}$ would not qualify, since it is undefined at $x = 0$.) Multiplying both sides by the same nonzero constant is undone by dividing by it, and dividing by a nonzero constant is undone by multiplying. The word nonzero is the whole point: multiplying both sides by $0$ turns every equation into $0 = 0$, which is true for all $x$. The new equation $0 = 0$ keeps no record of where you started, so the step cannot be reversed.

The trouble starts when the thing you multiply or divide by is not a constant but an expression that can equal zero. Then the step is reversible for most values of $x$ but not all, and the exceptions are exactly the solutions that get gained or lost.

Multiplying both sides by an expression that can be zero can add solutions. Multiply both sides of an equation by $x - 2$. Wherever $x - 2 \neq 0$ the step is reversible, since you can divide by $x - 2$ to undo it, so those solutions stay safe. But at $x = 2$ the multiplier is zero, and the new equation reads $0 = 0$ there whether or not $x = 2$ solved the original. So $x = 2$ can slip into the new solution set without belonging to the old one. A value that solves the new equation but not the original is called extraneous, and it is always found among the zeros of whatever you multiplied by.

Dividing both sides by an expression that can be zero can destroy solutions. Consider $x^2 = x$. Its solution set is $\{0, 1\}$, because $x^2 = x$ means $x^2 - x = 0$, that is $x(x - 1) = 0$, which holds exactly when $x = 0$ or $x = 1$. Now divide both sides of $x^2 = x$ by $x$. You get $x = 1$, whose solution set is just $\{1\}$. The solution $x = 0$ has vanished, and no wonder: dividing by $x$ is legal only when $x \neq 0$, so the step quietly threw away the very value it could not handle. A solution removed by an illegal step is a lost solution.

A reversible step always preserves the solution set exactly, which is the guarantee just proved. A step you cannot undo is not automatically wrong; it is merely not guaranteed to be safe. That missing guarantee is why a step that might add extra solutions needs its candidates checked against the original equation at the end rather than trusted. A step that might lose a solution needs a different fix, since a lost solution never shows up as a candidate to check in the first place. When a step you cannot undo does change the solution set, there are only two ways the set can change, and they are mirror images. Multiplying by something that is zero at a point, where the inverse fails, can add an extraneous solution there. Dividing by something that is zero at a point, where the step itself is illegal, can lose a solution there. Added roots and lost roots are the only changes a step can make, so every safe-solving rule you meet is really a way of ruling one of them out.

Every linear equation becomes ax = b

A linear equation in one variable is one where the variable appears only to the first power on each side, never squared, never multiplied by itself, and never hidden inside a denominator or a root. Two different kinds of safe move turn every linear equation into the single form $ax = b$. Distributing across parentheses and combining like terms are rewrites of a single side: they replace an expression by an equal one, justified by the distributive law and by arithmetic. Both rewrites are safe because after the rewrite the two sides still name the same number for every $x$, so the solution set cannot change. Moving a variable term or a constant to the other side is the second kind. That move adds the same expression to both sides, a reversible operation of the sort proved safe above. Both kinds preserve the solution set, for these two different reasons. Whatever tangle you start with, they collect it into

where $a$ and $b$ are specific real numbers. This single form settles the equation completely, and there are exactly three cases according to whether $a$ is zero.

Reducing to $ax = b$ is not busywork. It is the whole classification packed into two letters. The next lesson lets $a$ and $b$ be built from other letters, solving a formula for one of its variables. The only real question is then which of these three cases you land in as those letters change.

Clearing denominators changes the domain

Equations with the variable in a denominator are solved by multiplying both sides by the least common denominator to clear the fractions. That is convenient, but look at what kind of move it is. The denominators can be zero, so multiplying by them is the very “multiply by an expression that can be zero” step that adds extraneous solutions. Clearing denominators is the single most common place extraneous roots appear.

The fix is to respect the domain of the original equation. A fraction is undefined when its denominator is zero, so any value that makes a denominator zero is excluded. An excluded value is not eligible to be a solution, no matter what the algebra says later. Excluded is not the same as failing: a value the equation is not even defined at is not tested at all, so it cannot make the equation true or false. This is where the domain from the three cases stops being all of $\mathbb{R}$ by default. An equation whose domain skips a point can still be an identity on that domain, as long as it holds at every value the domain does include. The discipline is short. First, before clearing anything, list the excluded values. Then clear the denominators and solve the resulting equation. Finally, check each candidate against the excluded list and throw out any that appears there, because those are extraneous.

Return to the equation that opened the lesson, $\frac{x}{x - 2} = \frac{2}{x - 2}$. The denominator $x - 2$ is zero at $x = 2$, so $x = 2$ is excluded from the domain before we touch the algebra. Multiplying both sides by $x - 2$ gives $x = 2$, the one candidate. But $2$ is exactly the excluded value, so it is extraneous and gets discarded. Nothing survives, and the solution set is $\varnothing$. The equation genuinely has no solution: for every $x$ where the two sides are even defined, the only way they could be equal forces $x = 2$, which is not allowed. That clean step $x = 2$ was true only about the cleared equation, not the original, and the domain is what told them apart.

Common mistakes