Question

solve the system of linear equation by gauss elimination method 2x+3y-z=5;4x+4y-3z=3;2x-3y+2z=2

Answer 1

The solution for the linear equation for x,y,z is 1,2,3

Step-by-step explanation:

Given: $2x+3y-z=5$,

$4x+4y-3z=3$

$2x-3y+2z=2$

To find: we have to find the values of x,y,z using gauss elimination method.

Formula used: Ax=B is the system of linear equation

The linear equation is $2x+3y-z=5$,      (1)

                                     $4x+4y-3z=3$   (2)

                                     $4x+4y-3z=3$   (3)

First we have to solve for Z

Take the equation  2x+3y-z=5

$z=2x+3y-5$

Substitute $2x+3y-5$ for z in the (2)  and (3) equation

$4x+4y-3(2x+3y-5)=3$  

$2x-3y+2(2x+3y-5)=2$

Now we have to solve the above equation for x and y respectively

$4x+4y-6x-9y+15=3$

$-2x-5y+15=3$

rewrite the above equation as

$2x+5y=-3+15$

$2x+5y=12$

$5y=12-2x$

$y=\frac{12}{5} -\frac{2}{5} x$

 $2x-3y+4x+6y-10=2$

$6x+3y=12$

$6x=12-3y$

$x=\frac{12}{6} -\frac{3}{6} y$

$x=2-\frac{1}{2} y$

Substitute  $y=\frac{12}{5} -\frac{2}{5} x$ for y in the equation $x=2-\frac{1}{2} y$

$x=2-\frac{1}{2} (\frac{12}{5} -\frac{2}{5} x)$

$x=2-(\frac{12}{10}-\frac{2}{10} x)$

$x=2-(\frac{6}{5} -\frac{1}{5} x)$

$x=\frac{10-6}{5} +\frac{1}{5} x$

$x=\frac{4}{5} +\frac{1}{5}x$

$x-\frac{1}{5} x=\frac{4}{5}$

$4x=4$

$x=1$

Sub $x=1$ in the $y=\frac{12}{5} -\frac{2}{5} x$

$y=\frac{12}{5} -\frac{2}{5}$

$y=2$

Sub $x=1,y=2$ in the equation $z=2x+3y-5$

$z=2(1)+3(2)-5$

$z=2+6-5$

$z=8-5$

$z=3$

Therefore the solution for the linear equation is $x=1,y=2,$ and $z=3$