Question

find it quickly plz!!!!!!!!!!!!!!!!!!!!!!!!!!​

Answer 1

x+2y+3z = 14

3x + y + 2z = 11 ---------(2)

2x + 3y +z = 11 -----------(3)

multiplying eqn (2) by 2 and subtracting eqn(1) from it we get,

= 5x +2=8-------------(4)

again multiplying eqn (2) by 3 and

subtracting eqn (3) from it we get,

= 7x + 5z = 22 ------------(5)

Now multiply eqn (4) by 5 and subtract eqn (5) from it we get

18x = 18

..x=1

substituting the x in eqn (4) we get the value of z as

= 5(1)+2=8

:. z= 8 – 5 = 3

and substitute x and z in eqn (1) we get

1+2y+3(3) = 14

2y=14-1-9=4

..y=2

Answer 2

x+2y+3z = 14

3x + y + 2z = 11 ---------(2)

2x + 3y +z = 11 -----------(3)

multiplying eqn (2) by 2 and subtracting eqn(1) from it we get,

= 5x +2=8-------------(4)

again multiplying eqn (2) by 3 and

subtracting eqn (3) from it we get,

= 7x + 5z = 22 ------------(5)

Now multiply eqn (4) by 5 and subtract eqn (5) from it we get

18x = 18

..x=1

substituting the x in eqn (4) we get the value of z as

= 5(1)+2=8

:. z= 8 – 5 = 3

and substitute x and z in eqn (1) we get

1+2y+3(3) = 14

2y=14-1-9=4

..y=2

Answer 3

x+2y+3z = 14

3x + y + 2z = 11 ---------(2)

2x + 3y +z = 11 -----------(3)

multiplying eqn (2) by 2 and subtracting eqn(1) from it we get,

= 5x +2=8-------------(4)

again multiplying eqn (2) by 3 and

subtracting eqn (3) from it we get,

= 7x + 5z = 22 ------------(5)

Now multiply eqn (4) by 5 and subtract eqn (5) from it we get

18x = 18

..x=1

substituting the x in eqn (4) we get the value of z as

= 5(1)+2=8

:. z= 8 – 5 = 3

and substitute x and z in eqn (1) we get

1+2y+3(3) = 14

2y=14-1-9=4

..y=2

Answer 4

x+2y+3z = 14

3x + y + 2z = 11 ---------(2)

2x + 3y +z = 11 -----------(3)

multiplying eqn (2) by 2 and subtracting eqn(1) from it we get,

= 5x +2=8-------------(4)

again multiplying eqn (2) by 3 and

subtracting eqn (3) from it we get,

= 7x + 5z = 22 ------------(5)

Now multiply eqn (4) by 5 and subtract eqn (5) from it we get

18x = 18

..x=1

substituting the x in eqn (4) we get the value of z as

= 5(1)+2=8

:. z= 8 – 5 = 3

and substitute x and z in eqn (1) we get

1+2y+3(3) = 14

2y=14-1-9=4

..y=2