subject: Simultaneous Conditioning [print this page] In mathematics, the theory of linear systems is a branch of linear algebra, a subject which is fundamental to modern mathematics. Computational algorithms for finding the solutions and that are an important part of numerical linear algebra, and such methods play a prominent role in engineering. A system of non-linear equations can often to be approximated by a linear system (see linearization), a helpful technique when making a mathematical model or computer simulation of a relatively complex system. In this article we shall discuss about the problems involved in simultaneous conditioning.
There are Two Methods of Solving Simultaneous Conditioning :
Simultaneous equations, as the name suggests, are the two equations that are both true for a unique value of each of their variables. For example, the following two equations are both true for x = 1 and y = 2,
x + y = 3
x - y = -1
Elimination method
Substitution method
Problems for Simultaneous Conditioning
Elimination method to solve simultaneous equations :
This method can be used when coefficients of at least one of the variables is same in both equations, or if it can be easily made the same by multiplying either one or both the equations with a suitable number.
In equations where the coefficients of that two variables are the same, we use the elimination method, as it makes it easier to solve the equation,
For example,
2x + 4y = 14
2x + 3y = 8
In the above of the two equations, the coefficients of x in both the equations is the same, i.e. 2.
Thus we use the elimination method as follows:-
Subtract the two equations:
2x + 4y = 14
2x + 3y = 8
---------------
0 + y = 6
y = 6
Thus, y = 6. Now substitute the value of y = 6 in either one of the equations:-
2x + 4y = 14
2x + 4(6) = 14
2x + 24 = 14
2x = 14 - 24
2x = -10
x = '-10/5'
x = -2
Thus, x = -2
y = 6
In case the coefficient of the variables are different while multiplying the equations with a suitable number, we get similar coefficients, we can use the elimination method, as in
x + 3y = 7 - - - - - - (1)
2x + y = 4 - - - - - - (2)
On multiplying of the first equation by 2, we will get the same coefficients for x in both the equations,
Thus, y = 2. Substitute y = 2 in either of the two equations,
2x + y = 4
2x + 2 = 4
2x = 4 - 2
2x = 2
x = '2/2'
x = 1
Thus, x = 1
y = 2
Substitution method in simultaneous conditioning : -
When one of the equations is in the form in which one variable, either x or y, is isolated on one side of the equals sign, or when we see an equation that will be too difficult to be solved via elimination, the substitution method will make it easier to solve it.
In the substitution method, we can use one of the two equations to get the value of one variable in terms of the other variable. This value is not a number value, it is only an expression in terms of the other variable. 3x + 4y + 11 = 0; 5x + 6y = -7
3x + 4y + 11 = 0------------------(1)
5x + 6y = -7------------------------(2)
3x + 4y + 11 = 0
3x + 4y = -11
3x + = -11 -4y
x = ( -11 -4y) ------------------------------(3)
3
Substituting the value of x in equation number 2
Therefore, 5x + 6y = -7
5(-11 -'(4y)/3' ) + 6y = -7
-55 - '(20y)/3 ' + 6y = -7 ( multiplying both the sides by LCM 3)
