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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.

Two equations are equivalent on a specified domain of a function if they have the same solution set. Adding the same defined mathematical expression to both sides is reversible. Multiplying or dividing both sides by a mathematical expression is reversible only where that expression is nonzero and invertible in the arithmetic being used. In a field, every nonzero scalar is invertible; a general ring needs an additional check.

For example, $x^2=1$ and $(x-1)(x+1)=0$ are equivalent equations over the real numbers, by the distributive property. Concluding $x=1$ or $x=-1$ additionally uses the absence of zero divisors. The modular example in algebra shows why that extra inference can fail in a ring with zero divisors.

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.

For a function $f:X\to Y$ and $y\in Y$, its fiber of a function over $y$ is the set of all inputs producing that one output:Here $f^{-1}(y)$ is shorthand for a preimage of a singleton set, rather than a claim that an inverse function exists. A fiber can contain several inputs or be empty. It describes exactly which distinctions the function forgets.

For the finite function under image and preimage of a function, the fibers over $0,1,2,3$ are $\{a,b\}$, $\{c\}$, $\{d\}$ and the empty set, respectively. The nonempty fibers form a partition of a set: they are disjoint sets whose union is the domain of a function. Every input belongs to the fiber over its unique output, and two fibers with distinct output labels cannot share an input. A function is injective exactly when every fiber has at most one member; it is surjective exactly when every member of its specified codomain has a nonempty fiber.

More generally, $f^{-1}(U)$ is the union of the fibers over all $y\in U$. The relation $x\sim x^{\prime }$ defined by $f(x)=f(x^{\prime })$ is an equivalence relation, and its equivalence classes are precisely the nonempty fibers. The resulting quotient set is naturally in bijection with the image of a function. This links an input partition to the output values actually reached, without choosing one representative from each class.

Fibers are useful when an output is known but the input is not uniquely determined. A solution set $g^{-1}(\{0\})$ is one fiber, and factorization through a surjection explains when another function depends only on the retained output. Quantum hidden subgroup problem algorithms use functions whose fibers are cosets, giving the same idea a structured form in group theory.

A linear inequality compares a linear functional of a vector of unknowns with a constant, for example $a^{T}x\le b$ or $a^{T}x<b$ over the real numbers. For $a\ne 0$, its solution set is a closed or open half-space, respectively. The inequality $x+y\le 1$ allows all points on one side of the line $x+y=1$, including the line. Multiplication by a negative real number reverses the inequality, so $x\ge 1$ is equivalently $-x\le -1$.

Subtracting twice the first equation from the second gives $(\lambda +2)z=0$. The third equation gives $x=1+\lambda y$, and substitution in the first then gives $z=(\lambda +1)y$. Consequently $(\lambda +2)(\lambda +1)y=0$. The complete real solution set of this linear system isThere is no parameter value allowing both $x<0$ and $y<0$. In either exceptional family, $y=s<0$ forces $x>0$; in the remaining case, $y=0$. The exceptional values must be handled before dividing by their factors.

At $a=1$, the reduced system gives $z=1$ and $x+y=2$. Thus all solutions areGeometrically the solution set is an affine straight line, and substitution verifies all three original equations.

The original PDF has $\mathbf{c}=(-2,1,1)$ here; the transcription loses its minus sign. Now $\mathbf{a}\cdot \mathbf{b}=-1$ and $\mathbf{a}\cdot \mathbf{c}=0$, so the scalar condition becomes $0=0$ and leaves $s$ arbitrary. ConsequentlyThe solution set is a straight line parallel to $\mathbf{b}$, equivalently the intersection of $y=z$ and $x+y=-1$.

For the geometric mechanism, let $L=I+\mathbf{b}{\mathbf{a}}^T$. Since $\mathbf{a}\cdot \mathbf{b}=-1$, $L^2=L$, $L\mathbf{b}=0$, and $\mathbf{a}\cdot L\mathbf{r}=0$. Thus $L$ is an oblique projection onto the plane ${\mathbf{a}}^{\perp }$ along $\mathbf{b}$. The specified $\mathbf{c}$ lies in this image plane, and adding any multiple of the kernel direction $\mathbf{b}$ leaves its image unchanged.

The homogeneous linear system has solution set $\ker A$, a vector subspace. If it contains only the zero vector, no two different solutions of the full system can exist: their difference would be a nonzero homogeneous solution.

A particular solution ${\mathbf{v}}_p$ to the full system exists exactly when $\mathbf{w}$ belongs to the image of a linear map $\mathrm{im}A$. If it exists, every solution isIndeed, subtracting $A{\mathbf{v}}_p=\mathbf{w}$ from $A\mathbf{v}=\mathbf{w}$ gives $A(\mathbf{v}-{\mathbf{v}}_p)=0$, and conversely adding a kernel vector preserves the right-hand side. Thus there are no solutions, one solution if the kernel is trivial, or infinitely many solutions if it is nontrivial. The last solution set is an affine subspace of dimension $\dim \ker A$, determined by the rank-nullity theorem. For a square nonsingular matrix there is one solution for every right-hand side.

The complex modulus is a nonnegative real number. Equality therefore requires $z-1$ itself to be real and nonnegative. Its imaginary part gives $y=0$, and its real part then gives $x-1\ge 0$. Conversely every such point satisfies the equality. Hence the complete solution set isGeometrically this is the closed ray on the real axis starting at $z=1$.

For a function $f:X\to Y$ and a subset $U\subseteq Y$, its preimage isIt is a subset of the original domain of a function. The notation does not require an inverse function: the preimage of a singleton set can have no members, one member, or many members. In the finite example, $f^{-1}(\{0\})=\{a,b\}$, $f^{-1}(\{1\})=\{c\}$ and $f^{-1}(\{3\})=\varnothing$. The preimage of a singleton set is also its fiber of a function.

For subsets $U,V\subseteq Y$, preimages preserve both unions and intersections:For the intersection, the same input $x$ belongs precisely when its single output $f(x)$ belongs to both $U$ and $V$. There is no opportunity to choose different witnesses, which is what made the image counterexample fail. The identities also hold for arbitrary families of output subsets.

In the finite example, choose $U=\{0,1\}$ and $V=\{0,2\}$. ThenThe intersection of the two preimages is exactly the same set $\{a,b\}$. Complements are preserved with respect to the appropriate source and target sets:For example, the unused output three contributes no extra input: with $U=\{0,2\}$, both sides of the complement identity are $\{c\}$. See Lebl, Proposition 0.3.15: preimages of unions, intersections and complements.

This makes preimages useful for transporting conditions. The solution set of $g(x)=0$ is $g^{-1}(\{0\})$, and a zero set is a preimage of zero. Requiring two output conditions takes the intersection of their preimages. In topology, a continuous function pulls every open set back to an open set; a measurable function pulls the specified measurable sets in its codomain back to measurable sets in its domain of a function. These definitions use preimages because their set operations remain consistent even when the function is noninjective.

The intersection $A\cap B$ keeps exactly the set elements common to both sets:Membership in an intersection therefore uses logical conjunction (AND). For $A=\{1,2,3\}$ and $B=\{3,4\}$, the only shared member is $3$, so $A\cap B=\{3\}$. The Venn diagram in set union shades only the overlap for this operation.

An intersection is useful when every requirement must hold. If $A$ and $B$ are the sets of database records passing two separate conditions, their intersection gives the records passing both. Similarly, satisfying two equations requires belonging to the intersection of their solution sets. The same membership rule applies even when the objects have entirely different meanings.

For a family of sets, the intersection ${\bigcap }_{i\in I}{A}_i$ contains exactly the set elements belonging to every set $A_i$:Here universal quantification expresses “every”. If all the sets lie within a fixed universal set $U$, the intersection over no indexing members is defined to be $U$: there are no requirements to fail. This convention depends on specifying $U$. By contrast, intersecting with the empty set always gives $A\cap \varnothing =\varnothing$.

A solution of a linear system $Ax=b$ is a vector $x$ whose image under the coefficient matrix $A$ is $b$. Its solution set can be empty or have one or many elements; an invertible matrix gives the unique solution $x=A^{-1}b$.

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.

A system of linear inequalities requires several linear inequalities to hold simultaneously. The notation $Ax\le b$ compares coordinates componentwise, giving one inequality per row of the matrix $A$. Its solution set is the intersection of the corresponding half-spaces. For non-strict inequalities it is a linear polyhedron, which may be empty. For example, $x\ge 1$, $y\ge 1$, $x+y\le 1$ are inconsistent: adding their equivalent forms $-x\le -1$, $-y\le -1$, $x+y\le 1$ gives $0\le -1$. A Farkas certificate for linear inequalities generalizes this cancellation.