Equation solving

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An example of using Newton-Raphson method to solve the equation f(x)=0 or equivalently, to find a root of f (when f is the depicted function). The Newton-Raphson method is a procedure to find a numeric solution.
The quadratic formula, the symbolic solution for the quadratic equation. By instantiating it with the coefficients and evaluating, the numeric solution for the quadratic formula with those coefficients is found.

In mathematics, to solve an equation is to find what values (numbers, functions, sets, etc.) fulfill a condition stated in the form of an equation (two expressions related by equality). When searching a solution, one or more free variables are designated as unknowns. A solution is an assignment of expressions to the unknown variables that makes the equality in the equation true. In other words, a solution is an expression or a collection of expressions (one for each unknown) such that, when substituted for the unknowns, the equation becomes an identity. A problem of solving an equation may be numeric or symbolic. Solving an equation numerically means that only numbers represented literately (not as a combination of variables), are admitted as solutions. Solving an equation symbolically means that expressions that may contain known variables or possibly also variables not in the original equation are admitted as solutions.

For example, the equation x + y = 2x – 1 is solved for the unknown x by the solution x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement. It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1. Or x and y can both be treated as unknowns, and then there are many solutions to the equation. (x, y) = (a + 1, a) is a symbolic solution. Instantiating a symbolic solution with specific numbers always gives a numerical solution; for example, a = 0 gives (x, y) = (1, 0) (that is, x = 1 and y = 0) and a = 1 gives (x, y) = (2, 1). Note that the distinction between known variables and unknown variables is made in the statement of the problem, rather than the equation. However, in some areas of mathematics the convention is to reserve some variables as known and others as unknown. When writing polynomials, the coefficients are usually taken to be known and the indeterminates to be unknown, but depending on the problem, all variables may assume either role.

Depending on the problem, the task may be to find any solution (finding a single solution is enough) or all solutions. The set of all solutions is called the solution set. In the example above, the solution (x, y) = (a + 1, a) is also a parametrization of the solution set with the parameter being a. It is also possible that the task is to find a solution, among possibly many, that is best in some respect; problems of that nature are called optimization problems; solving an optimization problem is generally not referred to as "equation solving".

A wording such as "an equation in x and y", or "solve for x and y", implies that the unknowns are as indicated: in these cases x and y.


In one general case, we have a situation such as

ƒ (x1,...,xn) = c,

where x1,...,xn are the unknowns, and c is a constant. Its solutions are the members of the inverse image

ƒ −1[c] = {(a1,...,an) ∈ T1×···×Tn | ƒ (a1,...,an) = c},

where T1×···×Tn is the domain of the function ƒ. Note that the set of solutions can be the empty set (there are no solutions), a singleton (there is exactly one solution), finite, or infinite (there are infinitely many solutions).

For example, an equation such as

3x + 2y = 21z

with unknowns x, y and z, can be solved by first modifying the equation in some way while keeping it equivalent, such as subtracting 21z from both sides of the equation to obtain

3x + 2y − 21z = 0

In this particular case there is not just one solution to this equation, but an infinite set of solutions, which can be written

{(xyz) | 3x + 2y − 21z = 0}.

One particular solution is x = 0, y = 0, z = 0. Two other solutions are x = 3, y = 6, z = 1, and x = 8, y = 9, z = 2. In fact, this particular set of solutions describes a plane in three-dimensional space, which passes through the three points with these coordinates.

Solution sets[edit]

Main article: Solution set

A solution set is the set of values that satisfy a given set of equations or inequalities.

If the solution set is empty, then there are no xi such that the equation


in which c is a given constant, becomes true.

For example, let us examine a classic one-variable case. Using the squaring function on the integers, that is, the function ƒ whose domain are the integers (the whole numbers) defined by:

ƒ (x) = x2,

consider the equation

ƒ (x) = 2.

Its solution set is {}, the empty set, since 2 is not the square of an integer, so no integer solves this equation. However note that in attempting to find solutions for this equation, if we modify the function's definition – more specifically, the function's domain, we can find solutions to this equation. So, if we were instead to define that the domain of ƒ consists of the real numbers, the equation above has two solutions, and its solution set is

{√2, −√2}.

We have already seen that certain solutions sets can describe surfaces. For example, in studying elementary mathematics, one knows that the solution set of an equation in the form ax + by = c with ab, and c real-valued constants, with a and b not both equal to zero, forms a line in the vector space R2. However, it may not always be easy to graphically depict solutions sets – for example, the solution set to an equation in the form ax + by + cz + dw = k (with a, b, c, d, and k real-valued constants) is a hyperplane.

Methods of solution[edit]




Brute force, trial and error, inspired guess[edit]



Elementary algebra[edit]


8x+7=4x+35  \quad \text{or} \quad \frac{4x + 9}{3x + 4} = 2 \, ,

can be solved using the methods of elementary algebra.

Systems of linear equations[edit]


Polynomial equations[edit]


4x5x3 − 3 = 0

(by using the rational root theorem), and

x6 − 5x3 + 6 = 0,

(by using the substitution x = z1/3, which simplifies this to a quadratic equation in z).

Diophantine equations[edit]





Inverse functions[edit]

See also: Inverse problem


h(x) = c, c constant

by considering what is known as the inverse function of h.

Given a function h : AB, the inverse function, denoted h−1, defined as h−1 : BA is a function such that

h−1(h(x)) = h(h−1(x)) = x.

Now, if we apply the inverse function to both sides of


we obtain

h−1(h(x)) = h−1(c)
x = h−1(c)



h(h−1(x)) = x

holds. For example, the projection π1 : R2R defined by π1(x, y) = x has no post-inverse, but it has a pre-inverse π1−1 defined by π1−1(x) = (x, 0). Indeed, the equation

π1(x, y) = c

is solved by

(x, y) = π1−1(c) = (c, 0).




\tan x + \cot x = 2


\frac{\tan^2 x  -2 \tan x+1}{\tan x} = 0,

which can be factorized into

\frac{(\tan x - 1)^2}{\tan x}= 0.


x = \tfrac{\pi}{4} + k\pi, k = \cdots, -2, -1, 0, 1, 2, \ldots.

Numerical methods[edit]


Taylor series[edit]

One well-studied area of mathematics involves examining whether we can create some simple function to approximate a more complex equation near a given point. In fact, polynomials in one or several variables can be used to approximate functions in this way – these are known as Taylor series.

Matrix equations[edit]


Differential equations[edit]


See also[edit]