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In the simple case of a function of one variable, say, h(x), we can solve an equation of the form h(x) = c for some constant c by considering what is known as the inverse function of h. Given a function h : A → B, the inverse function, denoted h −1 and defined as h −1 : B → A, is a function such that
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
The unique pair of values a, b satisfying the first two equations is (a, b) = (1, 1); since these values also satisfy the third equation, there do in fact exist a, b such that a times the original first equation plus b times the original second equation equals the original third equation; we conclude that the third equation is linearly ...
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.
In elementary algebra, parentheses ( ) are used to specify the order of operations. [1] Terms inside the bracket are evaluated first; hence 2×(3 + 4) is 14, 20 ÷ (5(1 + 1)) is 2 and (2×3) + 4 is 10. This notation is extended to cover more general algebra involving variables: for example (x + y) × (x − y). Square brackets are also often ...
This counterintuitive result occurs because in the case where =, multiplying both sides by multiplies both sides by zero, and so necessarily produces a true equation just as in the first example. In general, whenever we multiply both sides of an equation by an expression involving variables, we introduce extraneous solutions wherever that ...
For example, consider the ordinary differential equation ′ = + The Euler method for solving this equation uses the finite difference quotient (+) ′ to approximate the differential equation by first substituting it for u'(x) then applying a little algebra (multiplying both sides by h, and then adding u(x) to both sides) to get (+) + (() +).
The mass of some objects on the scale is unknown and represents variables. Solving an equation corresponds to adding and removing objects on both sides in such a way that the sides stay in balance until the only object remaining on one side is the object of unknown mass. [145]