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+ X2 = 10 in .NET Printing PDF417 in .NET + X2 = 10

+ X2 = 10 use none none integrating tomake none with none ISO Standards Overview which has the unique solution equations such as = 3,X2 = 1. We will look at systems of language X I =EUa X I Ub X2 none none X 2 = 0u b X I u0 X2 which has the (unique) solution Xl=(aUbb)*,X2=b (aUbb)*. Checking that this is a solution entails verifying that both equations are satisifed if these expressions are substituted for the variables Xl and X 2. The solution of such systems parallels the solution of algebraic equations.

For example, the system. Regular Expressions Chap. 6 3Xl + X2 = 10 Xl -X 2 = 2 can be solved by treating the second statement as an equation in just the variable X2 and solving as indicated by the algebraic theorem "For any numbers a and b, where a =1= 0, the equation ax = b has a unique solution given by X = b -:- a." The second statement can be written as (-1)x2 = 2 - X2, which then admits the solution X2 = (2 - Xl) -:- (-1) or X2 = Xl - 2. This solution can be inserted into the first equation to eliminate X2 and form an equation" solely in Xl" Terms can be regrouped and the algebraic theorem can be applied to find Xl.

We would have 3Xl + X2 = 10 which becomes or 4Xl-2= 10 or. or Xl = 12 -:- 4 yi none for none elding xl=3 This value of Xl can be back-substituted to find the unique solution for X2: X2 = Xl - 2 = 3 - 2 = 1. Essentially, the same technique can be applied to any two equations in two unknowns, and formulas can be developed that predict the coefficients for the reduced set of equations. Consider the generalized system of algebraic equations with unknowns Xl and X2, constant terms El and E 2, and coefficients An, A 12 , A 2b and A 22 : AnXl + A12X2 = El A 2lXl + A.

22X 2 =. Recall that the app none for none ropriate formulas for reducing this to a single equation of the form .Anxl = E l , where the new coefficients.A n and El can be calculated as El = ElA22 - E2Al2 .

An = An A 22 - A12A2l A similar technique can be used to eliminate variables when there is a larger number of equations in the system. The following theorem makes similar predictions of the new coefficients for language equations..

Sec. 6.3 . Language Equations V Theorem 6.2. Let n 2= 2 and consider the system of equations in the unknowns Xl, Xz, .

.. ,Xn given by.

E1 U A ll X 1 U A 12XZU ... U A 1 (n-1)Xn- 1U A 1n X n Xz = Ez U A Z1 X1U AzzXz U ... U A Z(n-1)X n- 1U AznXn X n- 1= E n- 1U A(n-1)lX1 U A(n-1}2Xz U ... U A(n-1)(n-1)Xn- 1U A(n-1)n Xn Xn = En U A n1 X 1U AnzXz U ... U A n(n-1)Xn- 1U AnnXn in which (Vi,j E {I none none , 2, ...

,n})(X. $. Aii)" a.

This system has a unique solution. b. Define 13i = Ei U (An A:n .

En) for all i and. Ai = Aii U (Ain A:n . Ani) for all i,j = 1,2, ..

. ,n - 1..

= 1,2, ...

,n -. The solution of the original set of equations will agree with the solution of the following set of n - 1 equations in the unknowns Xl, Xz, ...

,Xn - 1 :. 131 U All x1 U A12x none none Z U ...

U A 1 (n-1)X n- 1. Xz = 13Z U A Z1 X 1 U Azzxz U ...

U A Z (n-1)Xn- 1. X n- 1= 13n- 1U A(n none none - 1)lX1 U A(n-1)ZXZU ...

U A(n-1)(n-1)Xn- 1. c. Once the solutio n to the above n - 1 equations in (b) is known, that solution can be used to find the remaining unknown:. = A:n (En U A n1 X none for none 1U AnzXz U ...

U A n(n-1)X n- 1). Proof. The proof hi nges on the repeated application of Theorem 6.1.

The last of the n equations, Xn = En U A n1 X 1U AnzXz U ...

U A n(n-1)Xn- 1U AnnXn can be thought of as an equation in the one unknown Xn with a coefficient of Ann for X n, and the remainder of the expression a "constant" term not involving X n The following parenthetical grouping illustrates this viewpoint:.
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