Solve the following system of equations 2x 3y 21 x3y 4z 8

Solve the following system of equations.

2x -3y + 2=1

x-3y + 4z = 8

3x + y -2z =2

There is one solution. The solution is (_, _, _)

Please explain the steps.

Solution

2x -3y + 2z =1       (1)         \"is this \"2\" or \"2z\" , considering 2z if anything else you can comment and i will solve that\"

x-3y + 4z = 8          (2)

3x + y -2z =2         (3)

we can eliminate any variable from these equations

let consider we have to eliminate \"x\"

therefore eq(2) X 2 - eq(1) we get

eq(2) x 2 => 2x - 6y +8z =16

eq(2) X 2 - eq(1)

2x   - 6y +8z =16

2x     -3y +2z = 1

-        +     -       -         // subtracting we reverse the sign

-3y + 6z =15           (4)

similarly eq(2) x 3 - eq(3)

3x   - 9y   +12z = 24

3x    +y   -2z      = 2

-       -         +         -

-10y +14z = 22          (5)

now we have eq(4) and (5) with only y and z terms

now we can eliminate any one \"y or z\"

eq(4) x10   - eq(5) X 3

-30y  + 60z = 150

-30y   +42z = 66

+        -          -

18z = 84

z = 84/18

z = 42/9

z= 14/3

suntittuiting this value in (4) or (5) to get the value of y

therefore from eq(4)

-3y + 6z =15

-3y +6(14/3) = 15

-3y + 28 =15

-3y = -13

y = 13/3

now we can find the value for \"x\" by using value of \"y\" and \"z\" from eq(1)or eq(2) or eq(3)

considering eq(1)

2x -3y +2z = 1

2x -3(13/3) +2 (14/3) =1

2x -13 +28/3 =1

2x = 1 + 13 - 28/3

2x = 14 -28/3

2x = (42-28)/3

2x = 14/3

x= 7/3

therefore

x=7/3 , y=13/3 and z= 14/3