Question

2x +3y+z=9,x+2y+3z=6,3x+y+2z=8 solve by gauss elimination method​

Answer 1

Answer:

The answers are:

X=35/18

Y=29/18

Z=5/18

Answer 2

Given Data:  $2x +3y+z=9, x+2y+3z=6, 3x+y+2z=8.$

To Find: Solve by Guass Elimination Method.

Solution:

• Mark the equations As 1, 2 and 3 etc.    

$2x +3y+z=9$ ------ ( 1 ),   $x+2y+3z=6$ ------ ( 2 ),  $3x+y+2z=8$ ------ ( 3 ).

• Multiplying -2 with (2) we get, $-2x-4y-+6z = -12$.

• Adding ( 4 ) with ( 1 ) we get,  

                                       $-2x-4y-6z = -12\\+\\2x +3y+z=9$   ;  $0-y-5z =-3$ -----( 4 ).

• Multiplying -3 with ( 2 ) we get, $-3x-6y-9z = -18$. Add ( 3 ) to it.

                                     $-3x-6y-9z = -18\\+\\3x+y+2z=8$; $0-5y-7z=-10$ ----- ( 5 ).

• By Multiplying -5 with ( 4 ) we get z, $5y+25z=15$  ; Add it with ( 5 ).

                                      $5y+25z=15\\+\\-5y-7z=-10$   ;    $18z = 5$

                                       $z = \frac{5}{18}$.

• By substituting z value in ( 4 ) we get y, $-y - 5(\frac{5}{18} ) = -3$.

                                     $y=3-\frac{25}{18}$.             ;                 $y= \frac{54-25}{18}$

                                        $y=\frac{29}{18}$.

• By substituting y and z in ( 2 ) we get x,

                                    $x+ 2(\frac{29}{18} ) + 3(\frac{5}{18} ) = 6 \\\\x+ \frac{58}{18}+\frac{15}{18} = \frac{108}{18} \\\\x = \frac{108-73}{18} \\\\x=\frac{35}{18}$

• Hence the Value of x, y, z is $\frac{35}{18}, \frac{29}{18}, \frac{5}{18}.$