Simultaneous Linear Equation

• Major: Chemical Engineering • Subject: Chemical Engineering Mathematics 2

•  Author: •  Andrew KUM KUMORO ORO

• e!t" o# Chemical Engineering • i!onegoro Uni$ersit% • 2&'(

CHAPTER 2 LINEAR EQUATIONS and  and SIMULTANEOUS LINEAR EQUATIONS

SUB CHAPTER Linear

equations

Simultaneous

linear equations with two

unknowns

Simultaneous

three unknowns

linear

equations

with

LINEAR EQUATIONS

Solution of simple equations A linear equation in a single variable (unknown) involves powers of the variable no higher than the rst. A linear equation is also referred to as a simple equation.  The solution of simple equations consists essentially of simplifying the expressions on each side of the equation to obtain an ax + b = cx + d giving ax − cx = d −b and hence equation of the form  x =

d −b a −c

SIMULTANEOUS LINEAR EQUATIONS WITH TWO UNKNOWNS

 Solution

by graphical methos

 Solution

by substitution

 

Solution by coecients/Elimination

equating

Simultaneous linear euations !it" t!o un#no!ns

Solution by graphical method  !et us consider the following system of two simultaneous linear equations in two variable.  "x # y $ %&  'x  "y $  *ere we assign any value to one of the two variables and then determine the value of the other variable from the given equation.

)or the e*uation  2+ ,% - .' .../'0

X

0

2

 2+ 1' - %

Y

1

5

X

3

-1

Y

0

6

 - 2+ 1 ' (+ 1 2% - 3 ... /20 2% - 3 , (+ 3 . (+  - ........... 2

Y

(-1,6)

(2,5)

(0,3)

(0,1)

X

X’ Y’

X= 1 Y=3

Simultaneous linear euations !it" t!o un#no!ns

Solution by substitution A linear equation in two variables has an innite number of solutions. +or two such equations there may be ,ust one pair of  x % and y %values that satisfy both simultaneously. +or(aexample ) 5x + 2 y =14 (b)

3x − 4 y = 24 from (a): 5 x + 2 y =14

 

∴ 2 y =14 −5 x ∴ y = 7 − 5 x

5 x 

in (b)

3x − 4  7 −

= 24 ∴ x = 4 ÷ 2 

in (a)

5(4) + 2 y =14

∴ y = −3

2

Simultaneous linear euations !it" t!o un#no!ns

Solution by equating coecients/Elimination -xample (a) (b)

3x + 2 y =16 4 x − 3 y =10

ultiply (a) by ' (the coe/cient of y  in (b)) and multiply (b) by " (the coe/cient of y  in (a)) (a) × 3 (b) × 2

9 x + 6 y = 48 8 x − 6 y = 20 add together to give 17 x = 68 ∴ x = 4

!"#tit!te in (a) to give 3(4) + 2 y =16

∴ y=2

Simultaneous linear euations !it" t"ree un#no!ns 0ith three unknowns and three equations the method of solution is ,ust an extension of the work with two unknowns. 1y equating the coe/cients of one of the variables it can be eliminated to give two equations in two unknowns. These can be solved in the usual manner and the value of the third variable evaluated by substitution.

Simultaneous linear euations Pre-simplifcation 2ometimes3 the given equations need to be simplied before the method of solution can be carried out. +or example3 to solve 2( x + 2 y) + 3(3x − y ) = 38 4(3 x + 2 y) − 3( x + 5 y) = −8

11 x + y = 38 2implication yields 9 x − 7 y = −8

atri! "orm o# Linear Equations $he #%#tem of e&!ation#:

+ a12T2 + ! + a1 N TN  = C1   a21T1 + a22T2 + ! + a2 N TN  = C2   a11T1





 

a N 1T1 + a N 2T2 + !



+ aNN TN = CN   

A total of 4 algebraic equations for the 4 nodal points and the system can be expressed as a matrix formulation

 a11

a 21  where A'   a   N 1

a12

!

a22

!

 aN 2

 !

a1 N  5A65T6$5C6  T1   C1

a2 N 

   a NN

T

 =   

T   C  2  C =  2     T   C  N   N 

     

Numeri$al Solutions

atrix form 5A65T6$5C6. +rom linear algebra 5A6%&5A65T6$5A6% & 5C63 5T6$5A6%&5C6 where 5A6%& is the inverse of matrix 5A6. 5T6 is the solution vector. atrix inversion requires cumbersome numerical computations and is not e/cient if the order of the matrix is

Numeri$al Solutions

9auss elimination method and other matrix solvers are usually available in many numerical solution package. +or example3 :4umerical ;ecipes< by =ambridge >niversity ?ress or their web source at www.nr.com. +or high order matrix3 iterative methods are usually more e/cient. The famous  @A=1B BT-;ATB4 C 9A>22%2-BD-! BT-;ATB4  methods will be introduced in the following.

Iteration "or Sol%in& Simulatenous Linear Equations

enera* a*ge"raic e&!ation for noda* +oint: i −1

N 

 j =1

j =i +1

∑ aijT j + aiiTi + ∑ aijT j = Ci   (,-am+*e : a31T1 + a32T2 + a33T3 + ! + a1 N TN  = C1  i = 3) .e/rite the e&!ation of the form: Ti

(k )

=

C i aii

i −1

−∑

aij

 j =1 aii

(k ) Tj

−

 N 

aij

∑a

j =i +1

( k −1)

;eplace (k) by (k%&) for the @acobi iteration

T j

ii

E (k) % specify the level of the iteration3 (k%&) means the present level and (k) represents the new level. E   An initial guess (k$8) is needed to start the iteration. E   1y substituting iterated values at (k%&) into the equation3 the new values at iteration (k) can be estimated

,X;, o*ve the fo**o/ing #%#tem of e&!ation# !#ing (a) the aco"i metho# (") the a!## eide* iteration method .eorganie into ne/ form: 4 X + 2Y + Z  = 11 11 1 1 X' - Y-  4 2 4 2 X + Y + 4 Z  = 16 3 1 Y '  X0  4 2 1  X   11 2 2  −1 2 0 Y  = 3      1 1 '4- X- Y  2 1 4  Z   16 2 4 (a) Jacobi method: !#e initia* g!e## X0'Y0'0'1 #to+ /hen maX -X-1Y -Y-1 --1 ≤ 01 First iteration: X1 ' (114) - (12)Y0 - (14)0 ' 2 Y1 ' (32)  (12)X 0 ' 2 1 ' 4 - (12) X0 - (14)Y0 ' 134 − X + 2Y + 0  Z  = 3

!#e the iterated va*!e# X 1'2 Y1'2 1'134 X2 ' (114) - (12)Y1 - (14)1 ' 1516 Y2 ' (32)  (12)X1 ' 52 2 ' 4 - (12) X 1 - (14)Y1 ' 52

 Second iteration:

Converging roce##: 13  15 5  5 7 63 93 133 31   393 A111B  22             4  16 2  2 8 32 32 128 16   128  519 517 767  512  256  256  to+ the iteration /hen ma-

 X

5

− X 4  Y 5 − Y 4  Z 5 − Z 4 ≤ 01

; ?;@$=?> A1014 202 2996B ,XC$ ?;@$=?> A1 2 3B

(b) $auss%Seidel iteration 2ubstitute the iterated values into the iterative process immediately after they are computed. @#e initia* g!e## X 0  X

=

11 4

−

1 2

Y

−

1 4

= Y 0 = Z 0 = 1

Z Y