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The solution set for the equations x − y = −1 and 3x + y = 9 is the single point (2, 3). A solution of a linear system is an assignment of values to the variables ,, …, such that each of the equations is satisfied. The set of all possible solutions is called the solution set. [5]
For example, antiderivatives of x 2 + 1 have the form 1 / 3 x 3 + x + c. For polynomials whose coefficients come from more abstract settings (for example, if the coefficients are integers modulo some prime number p , or elements of an arbitrary ring), the formula for the derivative can still be interpreted formally, with the coefficient ...
The system + =, + = has exactly one solution: x = 1, y = 2 The nonlinear system + =, + = has the two solutions (x, y) = (1, 0) and (x, y) = (0, 1), while + + =, + + =, + + = has an infinite number of solutions because the third equation is the first equation plus twice the second one and hence contains no independent information; thus any value of z can be chosen and values of x and y can be ...
For example, the polynomial x 2 y 2 + 3x 3 + 4y has degree 4, the same degree as the term x 2 y 2. However, a polynomial in variables x and y, is a polynomial in x with coefficients which are polynomials in y, and also a polynomial in y with coefficients which are polynomials in x. The polynomial
To do so, the different variables in the equation are understood as coordinates and the values that solve the equation are interpreted as points of a graph. For example, if x {\displaystyle x} is set to zero in the equation y = 0.5 x − 1 {\displaystyle y=0.5x-1} , then y {\displaystyle y} must be −1 for the equation to be true.
Horner's method can be used to convert between different positional numeral systems – in which case x is the base of the number system, and the a i coefficients are the digits of the base-x representation of a given number – and can also be used if x is a matrix, in which case the gain in computational efficiency is even greater.
If y 2 = x 3 − x − 1, then the field C(x, y) is an elliptic function field. The element x is not uniquely determined; the field can also be regarded, for instance, as an extension of C(y). The algebraic curve corresponding to the function field is simply the set of points (x, y) in C 2 satisfying y 2 = x 3 − x − 1.
To solve this kind of equation, the technique is add, subtract, multiply, or divide both sides of the equation by the same number in order to isolate the variable on one side of the equation. Once the variable is isolated, the other side of the equation is the value of the variable. [37] This problem and its solution are as follows: Solving for x