Answer:
Choice A is the correct answer.
Step-by-step explanation:
We have given a system of equations.
5x+2y+z = 4 eq(1)
x+2z = 4 eq(2)
2x+y-z = -1 eq(3)
We have to find the solution of the system.
From eq(2), we have
x = 4-2z eq(4)
Putting above eq(4) in eq(3), we have
2(4-2z)+y-z = -1
8-4z+y-z = -1
8-5z+y = -1
y-5z = -1-8
y-5z = -9 eq(a)
Multiplying above equation by 2 , we have
2y-10z = -18 eq(5)
Putting eq(4) into eq(1),we have
5(4-2z)+2y+z = 4
20-10z+2y+z = 4
2y-9z = -16 eq(6)
Subtracting eq(6) from eq(5), we have
2y-9z-(2y-10z) = -16-(-18)
2y-9z-2y+10z = -16+18
z = 2
Putting the value of z in eq(4), we have
x = 4-2(2) = 4-4
x = 0
Putting the value of z in eq(a), we have
y-5(2) = -9
y-10 = -9
y = -9+10
y = 1
Hence, the solution is x = 0,y = 1 and z = 2.
