Solve following system xyz3 2xyz2 3xyz5SolutionSolution Here

Solve following system:

x-y+z=3

2x+y+z=2

3x-y+z=5

Solution:

Here;

x-y+z=3 ---(1)

2x+y+z=2 -----(2)

3x-y+z=5 ------(3)

hence, (3)-(1):

3x-y+z - (x-y+z) =5-3

=> 2x = 2 => x = 1

Now, (1)+(2):

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

=> 3x+2z = 5

Putting the value of x =1

=> z = 1

Using these values x=1,z=1

by (1);

we get -

1-1+z = 3

=> z = 3

Hence the solution :

x = 1 , y= 1 and z =3