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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]
Vertical line of equation x = a Horizontal line of equation y = b. Each solution (x, y) of a linear equation + + = may be viewed as the Cartesian coordinates of a point in the Euclidean plane. With this interpretation, all solutions of the equation form a line, provided that a and b are not both zero. Conversely, every line is the set of all ...
[17] [18] For example, the fraction 1/(x 2 + 1) is not a polynomial, and it cannot be written as a finite sum of powers of the variable x. For polynomials in one variable, there is a notion of Euclidean division of polynomials, generalizing the Euclidean division of integers.
The formula is given in verses 17–19, chapter VII, Mahabhaskariya of Bhāskara I. A translation of the verses is given below: [3] (Now) I briefly state the rule (for finding the bhujaphala and the kotiphala, etc.) without making use of the Rsine-differences 225, etc. Subtract the degrees of a bhuja (or koti) from the degrees of a half circle (that is, 180 degrees).
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; a symbolic solution is (x, y) = (a + 1, a), where the variable a may take any value. Instantiating a symbolic solution with specific numbers ...
Once y is also eliminated from the third row, the result is a system of linear equations in triangular form, and so the first part of the algorithm is complete. From a computational point of view, it is faster to solve the variables in reverse order, a process known as back-substitution. One sees the solution is z = −1, y = 3, and x = 2. So ...
If two solutions intersect each other, that is, they both go through the same point (x,y), then there is a failure of uniqueness for a first-order ordinary differential equation. Thus, there will be a failure of uniqueness if a solution of the first form intersects the second solution. The condition of intersection is : y s (x) = y c (x). We solve
Therefore, let f(x) = g(x) = 2x + 1. Then, f(x)g(x) = 4x 2 + 4x + 1 = 1. Thus deg(f⋅g) = 0 which is not greater than the degrees of f and g (which each had degree 1). Since the norm function is not defined for the zero element of the ring, we consider the degree of the polynomial f(x) = 0 to also be undefined so that it follows the rules of a ...