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In the first step both numbers were divided by 10, which is a factor common to both 120 and 90. In the second step, they were divided by 3. The final result, 4 / 3 , is an irreducible fraction because 4 and 3 have no common factors other than 1.
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.
In mathematics, a rational function is any function that can be defined by a rational fraction, which is an algebraic fraction such that both the numerator and the denominator are polynomials. The coefficients of the polynomials need not be rational numbers ; they may be taken in any field K .
A rational algebraic expression (or rational expression) is an algebraic expression that can be written as a quotient of polynomials, such as x 2 + 4x + 4. An irrational algebraic expression is one that is not rational, such as √ x + 4.
In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.
If the expressions a and b are polynomials, the algebraic fraction is called a rational algebraic fraction [1] or simply rational fraction. [2] [3] Rational fractions are also known as rational expressions.
If the numerator and the denominator are polynomials, as in + , the algebraic fraction is called a rational fraction (or rational expression). An irrational fraction is one that is not rational, as, for example, one that contains the variable under a fractional exponent or root, as in x + 2 x 2 − 3 {\displaystyle {\frac {\sqrt {x+2 ...
Solving an equation symbolically means that expressions can be used for representing the solutions. 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.