Question

x + y + 2z = 1; 2x + y + z = 2; x + 2y + 2z = 1​

Answer 1

Subtracting eqn. (i) from (iii), we get,

y = 0

Now, eqn. (iii) by (ii), we get,

2x + 4y + 4z = 2 ...(iii)

since, y = 0

2x + 4z = 2 ...(iii)

and 2x + z = 2 ...(ii)

Therefore, subtracting (ii) from (iii), we get,

3z = 0

z = 0

Now, putting the values of z and y in eqn. (i), we get,

x + 0 + 0 = 1

=> x = 1

Ans: x=1, y=0 and z=0

Answer 2

Answer:

X=0 and Y = 0 and Z=½

Step-by-step explanation:

X+Y+2z = 1 1st equation

2x+ y+2 =2

=> 2x+y+2= 2(X+y+2z)

=> 2x+y+2= 2x + 2y +4z

=>4z+y=2 .....2nd equation

x+2y+2z = 1 = X+y+2z

=>y=2y

Thus, Y =0 Because only 2*0 = 0

Putting value of Y in 2nd equation

4z =2 => z=½

X+y+2z =1

=> X+1 = 1

=> X=0