Question

Solve the equation: x+2y+2z=11 2x+y+z=7, 3x+4y+z. ​

Answer 1

Answer:

let x+2y+3z=14 ------------ (1)

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

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

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

=5x+z=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)+z=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

Answer:

x=1, y=2, z=3

Step-by-step explanation:

let x+2y+3z=14 ------------ (1)

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

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

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

=5x+z=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)+z=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