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The simplified equation is not entirely equivalent to the original. For when we substitute y = 0 and z = 0 in the last equation, both sides simplify to 0, so we get 0 = 0, a mathematical truth. But the same substitution applied to the original equation results in x/6 + 0/0 = 1, which is mathematically meaningless.
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 ...
Divide the previously dropped/summed number by the leading coefficient of the divisor and place it on the row below (this doesn't need to be done if the leading coefficient is 1). In this case q 3 = a 7 b 4 {\displaystyle q_{3}={\dfrac {a_{7}}{b_{4}}}} , where the index 3 = 7 − 4 {\displaystyle 3=7-4} has been chosen by subtracting the index ...
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 ...
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 solutions –1 and 2 of the polynomial equation x 2 – x + 2 = 0 are the points where the graph of the quadratic function y = x 2 – x + 2 cuts the x-axis. In general, an algebraic equation or polynomial equation is an equation of the form =, or = [a]
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.
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]