Search results
Results From The WOW.Com Content Network
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]
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.
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
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 roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
The solutions of the corresponding equations get more complex. For equations of order three in the plane a fairly complete answer may be found in. [2] A typical example of a third-order equation that is also of historical interest is due to Blumberg. [14] Example 7 (Blumberg 1912) In his dissertation Blumberg considered the third order operator
y=f(x)=.5x+1 or f(x,y)=x-2y+2=0 Positive and negative half-planes. The slope-intercept form of a line is written as = = + where is the slope and is the y-intercept. Because this is a function of only , it can't represent a vertical line.
The result x 2 is then multiplied by the second term in the divisor −3 = −3x 2. Determine the partial remainder by subtracting −2x 2 − (−3x 2) = x 2. Mark −2x 2 as used and place the new remainder x 2 above it.