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An equation asserts that two mathematical expressions have the same value. Solving it means finding all allowed assignments to its variables that make that assertion true. It converts a relationship into a constraint: a measurement, a geometric intersection, or an equilibrium model can leave an unknown value to determine.

The allowed values matter. The equation $2x=1$ has solution set $\{1/2\}$ over the rational numbers, but the empty set over the integers. Over the real numbers, $x^2=1$ has two solutions, $-1$ and $1$. A mathematical identity, such as $(x+1{)}^2=x^2+2x+1$, holds for every permitted real $x$. The familiar equality sign therefore does not by itself say whether one is checking a particular value, solving for an unknown, or stating an identity.

For a worked linear equation over the real numbers,The first step subtracts three from both sides; the second divides both sides by the nonzero number two. Reversing the steps recovers the original equation, so these are equivalent equations with the same solution set $\{3\}$. Checking the value gives $2\cdot 3+3=9$. The role of reversible operations and the classification by solution sets are described in College Algebra 2e, §2.2: Solving Linear Equations in One Variable.

When the two sides define real-valued functions $f$ and $g$ on a common domain of a function, solutions are exactly the input values at which their function graphs intersect. Equivalently, they are the zeros of a function $f-g$. This gives a useful connection between symbolic algebra and geometry: rearranging an equation can simplify the comparison without changing the inputs where equality holds.

Not every operation is reversible on every domain of a function. Dividing $x(x-1)=0$ by $x$ gives $x-1=0$ only after restricting to $x\ne 0$; it loses the solution $x=0$ if that restriction is forgotten. Squaring both sides can instead introduce an extraneous solution, because distinct real numbers can have the same square. The source equation, its domain of a function, and the direction of each implication should travel together.

An extraneous solution satisfies a transformed equation but fails the original equation or its restrictions. The term describes a candidate generated during the calculation, rather than a true solution that can be kept in the final solution set.

Consider the real equationThe square root requires $x\ge 0$. Because the square root is nonnegative, any solution must also have $x-2\ge 0$, so $x\ge 2$. Squaring givesThe squared equation has candidates $x=1$ and $x=4$. At $x=4$, the original equality is $2=2$. At $x=1$, it is $1=-1$, which fails: this candidate is an extraneous solution. Thus the original solution set is exactly $\{4\}$.

On the restricted domain of a function $x\ge 2$, both original sides are nonnegative, so squaring is reversible there and the two equations are equivalent. Retaining the sign restriction or checking every candidate in the original equation achieves the same final solution set. The general need to check candidates after raising both sides to a power is explained in College Algebra 2e, §2.6: Solving Radical Equations.

The solution set of a mathematical condition $P$ in a specified domain $D$ collects every allowed value that satisfies it:It describes the answer without committing to the steps used to find it. For the equation $x^2=1$ over the real numbers, the solution set is $\{-1,1\}$; restricting to positive real numbers leaves $\{1\}$. For $2x=1$, the solution set is $\{1/2\}$ over the rational numbers and the empty set over the integers. The domain is part of the question.

An equation can have one solution, several solutions, no solutions, or an entire domain of solutions. Over the real numbers, $2x+3=9$ gives the singleton set $\{3\}$, $x^2+1=0$ gives the empty set, and the mathematical identity $(x+1{)}^2=x^2+2x+1$ gives the whole real line. Changing that second equation's domain to the complex numbers gives $\{-i,i\}$. The elementary definition by values that satisfy an equation and the classification by solution sets appear in College Algebra 2e, §2.2: Solving Linear Equations in One Variable.

Requiring several conditions simultaneously takes the intersection of their solution sets. The squared equation $x=(x-2{)}^2$ has solution set $\{1,4\}$. Adding the restriction $x\ge 2$ giveswhich is also the solution set of $\sqrt{x}=x-2$. The extraneous solution example explains why the sign restriction is needed. Requiring at least one of several conditions instead takes their union. This turns logical combinations of constraints into reusable operations on sets.

For several variables, a solution is an ordered assignment, usually represented by a tuple. The real equation $x+y=1$ has solution set $\{(t,1-t):t\in \mathbb{R}\}$, a straight line in the Euclidean plane with coordinates $(x,y)$. Adding $x-y=0$ intersects that straight line with another, leaving the single tuple $(1/2,1/2)$. A system of linear inequalities has no solution exactly when the intersection of its individual solution sets is the empty set. For optimization constraints, this common solution set is the feasible region.

Because a solution set is a set, the order in which its members are listed is irrelevant. The multiple root of $(x-1{)}^2=0$ is still the one solution $1$; its multiplicity of a root is additional polynomial information, rather than a second occurrence inside the solution set. Equivalent equations preserve the solution set on the stated domain, even if their algebraic forms differ.