Question

The solutions of the equations x+2y+3z=14, 3x+y+2z = 11, 2x + 3y + z = 11 ...

• A. x = 0, y = 2, z = 4
• B. x = 1, y = 0, z = 4
• C. x = 0, y = 1, z = 8
• D. x = 1, y = 2, z = 3

Answer 1

The correct option is D x = 1, y = 2, z = 3

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

$18x=18$

$\therefore x=1$

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

= $5\left(1\right)+z=8$

$\therefore z=8-5=3$

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

$1+2y+3\left(3\right)=14$

$2y=14-1-9=4$

$\therefore y=2$

hence the answer is option D