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Except for Bézout's theorem, the general approach was to eliminate variables for reducing the problem to a single equation in one variable. The case of linear equations was completely solved by Gaussian elimination , where the older method of Cramer's rule does not proceed by elimination, and works only when the number of equations equals the ...
An example from mathematics says that a single-variable quadratic polynomial has a real root if and only if its discriminant is non-negative: [1] ∃ x ∈ R . ( a ≠ 0 ∧ a x 2 + b x + c = 0 ) a ≠ 0 ∧ b 2 − 4 a c ≥ 0 {\displaystyle \exists x\in \mathbb {R} .(a\neq 0\wedge ax^{2}+bx+c=0)\ \ \Longleftrightarrow \ \ a\neq 0\wedge b^{2 ...
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 there is a unique solution to the original system of equations.
If the subexpressions are not identical, then it may still be possible to cancel them out partly. For example, in the simple equation 3 + 2y = 8y, both sides actually contain 2y (because 8y is the same as 2y + 6y). Therefore, the 2y on both sides can be cancelled out, leaving 3 = 6y, or y = 0.5. This is equivalent to subtracting 2y from both sides.
Since all the inequalities are in the same form (all less-than or all greater-than), we can examine the coefficient signs for each variable. Eliminating x would yield 2*2 = 4 inequalities on the remaining variables, and so would eliminating y. Eliminating z would yield only 3*1 = 3 inequalities so we use that instead.
One of the basic principles of algebra is that one can multiply both sides of an equation by the same expression without changing the equation's solutions. However, strictly speaking, this is not true, in that multiplication by certain expressions may introduce new solutions that were not present before. For example, consider the following ...
For stating the theorem in terms of commutative algebra, one has to consider a polynomial ring [] = [, …,] over a commutative Noetherian ring R, and a homogeneous ideal I generated by homogeneous polynomials, …,. (In the original proof by Macaulay, k was equal to n, and R was a polynomial ring over the integers, whose indeterminates were all the coefficients of the.
Difficult integrals may also be solved by simplifying the integral using a change of variables given by the corresponding Jacobian matrix and determinant. [1] Using the Jacobian determinant and the corresponding change of variable that it gives is the basis of coordinate systems such as polar, cylindrical, and spherical coordinate systems.