Linear equations

An equation states that two expressions are equal and contains an unknown. To solve an equation means to find every number that makes the statement true when substituted for the unknown.

An equation is linear when the unknown appears in the first power — not squared, not in a denominator, not under a root:

A balance scale, or equivalent equations

The equals sign works like a balance scale: both sides weigh the same. As long as we do exactly the same thing to both sides, the balance holds and the set of solutions does not change. Equations related that way are called equivalent.

Two operations preserve equivalence:

The condition $c \neq 0$ is not a formality: multiplying both sides by zero turns any equation into the true $0 = 0$, and dividing by zero cannot be done at all.

This is where the school phrase "move it across and change the sign" comes from. It is not a separate rule but shorthand: from $x + 5 = 12$ we subtract $5$ from both sides and are left with $x = 7$.

Solving step by step

The plan is always the same: tidy up both sides, gather the unknowns on one side and the numbers on the other, then divide by the coefficient of the unknown.

Checking is part of the solution, not decoration: it costs two multiplications and catches every lost sign.

The solution on a graph

This one section needs the coordinate plane, so what it needs is here. Two perpendicular axes — the horizontal $x$ and the vertical $y$ — meet at the point $(0, 0)$. Every point of the plane is then described by a pair of numbers: the first says how far to go right, the second how far up. So the point $(4, 11)$ sits four squares right of the $y$ axis and eleven up from the $x$ axis. That is everything the drawing below asks for; the coordinate plane itself — quadrants, distances, the midpoint of a segment — is a topic of its own: the coordinate plane.

The equation $2x + 3 = 11$ can be read as a question: for which $x$ does the line $y = 2x + 3$ reach the height $11$?

The lines cross at a single point, so the equation has exactly one solution. The same picture explains both special cases below: parallel lines never meet, while lines that coincide share every point.

When there is no solution — and when there are infinitely many

Sometimes the unknowns vanish along the way. What is left is a statement about numbers alone, and that statement decides:

An equation as a model

The power of equations comes from describing real situations. A taxi charges 8 for the ride plus 3 per kilometre; how far does 29 take us?

Seven kilometres. The hard part of such a task is rarely the arithmetic — it is writing the equation down.

Rearranging a formula

A formula is an equation in several letters: $s = vt$ ties distance, speed and time together, and $P = ab$ ties the area of a rectangle to its sides. It is always written with one of those letters as its subject, but the question is often about another one. Rather than substituting numbers and starting from scratch every time, it pays to make the wanted quantity the subject once — and get a ready-made formula out of it.

The rules are exactly the ones used with a single unknown. The letter we are after is the unknown; every other letter is treated as a number:

The condition $v \neq 0$ is not decoration. In an ordinary equation you can see what you are dividing by and check that it is not zero; with a letter nobody checks for you, so the condition has to be written down.

When the wanted letter appears in more than one place, gather it on one side first and factor it out — just as in algebraic expressions. From the perimeter of a rectangle $P = 2(a + b)$, the letter $a$ comes out in two moves:

An unknown in the denominator

The equation $\dfrac{12}{x + 1} = 4$ is not linear — the unknown sits in a denominator. One multiplication reduces it to a linear one, though, and from there the plan of this lesson takes over. The order of the steps is different from usual: the assumption comes before the arithmetic.

A denominator cannot be zero, because division by zero cannot be done. So the number that makes it zero is excluded:

The set of numbers for which the equation makes sense at all is its domain. For a linear equation $ax + b = c$ the domain is the whole of $\mathbb{R}$, which is why nobody mentions it. Here one number drops out of it — and it has to be written down before any rearranging, because multiplying by the denominator erases every trace of it.

The assumption, once written, is also the permission for the next step: since $x + 1 \neq 0$, both sides may be multiplied by that bracket.

That last step — comparing the result against the domain — is not a formality. Sometimes the only candidate is exactly the excluded number, and then the equation has no solutions, even though the arithmetic was flawless. So it goes with $\dfrac{x}{x - 3} = \dfrac{3}{x - 3}$: the domain is $x \neq 3$, multiplying by the denominator gives $x = 3$, and that number is not in the domain. The solution set is empty.

Equations with more fractions than one, or with a polynomial in the denominator, are the subject of rational expressions and equations. The scheme stays the same: domain, arithmetic, then check the candidates.

In the exercises below, enter the solution in the form $x = 5$ (a bare $5$ is accepted too). Where the task is to make a quantity the subject of a formula, the answer is an expression in the remaining letters — for $s = vt$ and the wanted $t$ that is $s/v$.

Common mistakes

• Moving a term across without changing its sign — $x + 5 = 12$ gives $x = 7$, not $x = 17$.
• Operating on one side only — when you divide, divide both sides.
• Losing a sign at a bracket — $-2(x - 5)$ is $-2x + 10$.
• Dividing by an expression containing the unknown — it may be zero, so solutions get lost.
• Dividing by a letter without a condition — $s = vt$ gives $t = \frac{s}{v}$ only when $v \neq 0$.
• Multiplying by the denominator before writing the domain — once multiplied, the excluded number is invisible.
• Accepting a candidate outside the domain — if the only candidate drops out of it, the equation has no solutions.
• Skipping the check — the one step that catches an arithmetic slip.