= Equation
{title2=$L=R$}
{wiki}
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>,
$$
2x+3=9
\quad\Longleftrightarrow\quad
2x=6
\quad\Longleftrightarrow\quad
x=3.
$$
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 $2\cdot3+3=9$. The role of reversible operations and the classification by <solution sets> are described in https://openstax.org/books/college-algebra-2e/pages/2-2-linear-equations-in-one-variable[College Algebra 2e, §2.2: Solving Linear Equations in One Variable].
\Image[/equivalent-linear-equation-intersections.png]
{title=Reversible operations change the graphs of the two sides but preserve the intersection's input value $x=3$}
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 $x\ne0$; 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.
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