To find the solution of the given system of linear equations:
x−y+z=4→(1) x−2y+2z=9→(2) 2x+y+3z=1→(3) Let's try to eliminate variables to solve for each variable and see which option fits:
First, subtract equation (1) from equation (2) to eliminate the variable x.
(x−2y+2z)−(x−y+z)=9−4 −y+z=5→(4) Now, we can use equations (4) and (1) to eliminate z : Multiply equation (4) by -1 and add to equation (1): z−y=−5 x−y+z=4 ⇒x+(z−y)+(y−z)=−1+4 x=−1→(5) After finding x, substitute x=−1 back into equation (4): −1−y+z=4 −y+z=5 ([Already derived as (4)]) And you already have the solution from (4): z=y+5 To find y, substitute equation (4) and x=−1 into equation (3): 2(−1)+y+3(z)=1 −2+y+3(y+5)=1 −2+y+3y+15=1 4y+13=1 4y=−12 y=−3→(6) Substitute y=−3 back into the relationship for z : z=−3+5 z=2→(7) Therefore, the solution to the system is: x=−1,y=−3,z=2 Thus, the correct answer is Option C: x=−1;y=−3;z=2