Question

X+4y+3z=17 , 3x+3y+z=16 , 2x+2y+z=11 find x y z

Answer 1

Step-by-step explanation:

x+4y+3z = 17............(1)

3x+3y+z= 16...........(2)

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

on multipliying by3 in (1) and substracting to (2)

3x+12y+9z= 51

3x+3y+z= 16

---------------------

9y+8z= 35.............(4)

on multipliying by 2 in (1) and substracting to (3)

2x+8y+6z= 34

2x+2y+z = 11

___________

6y+5z = 23...........(5)

=>eqn(4) ×5 - eqn(5) ×8

=> 45y+40z= 175

48y+ 40z= 184

________________ substracting

-3y= -9=> y=3

on putting y= 3 in eqn (4)

9×3+8z= 35

=> z= 1

on putting the value of y and z in (1)

x+ 4×3+3×1= 17

=> x= 2

therefore , x= 2, y= 3 and z= 1

again xyz = 2×3×1

xyz = 7