Search results
Results from the WOW.Com Content Network
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. It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1.
This shows again that any rational root of P is positive, and the only remaining candidates are 2 and 2\3. To show that 2 is not a root, it suffices to remark that if x = 2 , {\displaystyle x=2,} then 3 x 3 {\displaystyle 3x^{3}} and 5 x − 2 {\displaystyle 5x-2} are multiples of 8 , while − 5 x 2 {\displaystyle -5x^{2}} is not.
For example, the statement = is true if is either 2 or −2 and false otherwise. [26] Equations with variables can be divided into identity equations and conditional equations. Identity equations are true for all values that can be assigned to the variables, such as the equation 2 x + 5 x = 7 x {\displaystyle 2x+5x=7x} .
Problem in words: If you double the age of a child and add 4, the resulting answer is 12. How old is the child? Equivalent equation: 2 x + 4 = 12 {\displaystyle 2x+4=12} where x represent the child's age
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]
For example, the polynomial x 2 y 2 + 3x 3 + 4y has degree 4, the same degree as the term x 2 y 2. However, a polynomial in variables x and y, is a polynomial in x with coefficients which are polynomials in y, and also a polynomial in y with coefficients which are polynomials in x. The polynomial
This addition can multiply winnings by 2X, 3X, 4X, 5X, or 10X. Players can also ask a cashier for a "Quick Pick" where a cashier will give you computer generated numbers on a printed Powerball ticket.
Trigonometric identities may help simplify the answer. [ 1 ] [ 2 ] Like other methods of integration by substitution, when evaluating a definite integral, it may be simpler to completely deduce the antiderivative before applying the boundaries of integration.