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For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression 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.
For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0.
The equations 3x + 2y = 6 and 3x + 2y = 12 are inconsistent. A linear system is inconsistent if it has no solution, and otherwise, it is said to be consistent. [7] When the system is inconsistent, it is possible to derive a contradiction from the equations, that may always be rewritten as the statement 0 = 1. For example, the equations
Given two different points (x 1, y 1) and (x 2, y 2), there is exactly one line that passes through them. There are several ways to write a linear equation of this line. If x 1 ≠ x 2, the slope of the line is . Thus, a point-slope form is [3]
For example, the second-order equation y′′ = −y can be rewritten as two first-order equations: y′ = z and z′ = −y. In this section, we describe numerical methods for IVPs, and remark that boundary value problems (BVPs) require a different set of tools. In a BVP, one defines values, or components of the solution y at more than one ...
The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
For example, if a system contains , a system over the rational numbers is obtained by adding the equation r 2 2 – 2 = 0 and replacing by r 2 in the other equations. In the case of a finite field, the same transformation allows always supposing that the field k has a prime order.
Hilbert's tenth problem is the tenth on the list of mathematical problems that the German mathematician David Hilbert posed in 1900. It is the challenge to provide a general algorithm that, for any given Diophantine equation (a polynomial equation with integer coefficients and a finite number of unknowns), can decide whether the equation has a solution with all unknowns taking integer values.