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# Find all the solutions of the system of equations: x+2y-z=0, 2x+y+z=0, x-4y+5z=0.

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asked 2021-03-28
Find all the solutions of the system of equations:
x+2y-z=0, 2x+y+z=0, x-4y+5z=0.

## Answers (1)

2021-03-30
Step 1
we have to find all the solutions of the system of equations:
x+2y-z=0
2x+y+z=0
x-4y+5z=0
Step 2
the given system of equations is:
x+2y-z=0...(1)
2x+y+z=0...(2)
x-4y+5z=0...(3)
from the equation (1), we get
x+2y-z=0
x=-2y+z
now substitute the value of x in the equation (2) and (3),
2x+y+z=0
2(−2y+z)+y+z=0
−4y+2z+y+z=0
-3y+3z=0
-y+z=0...(4)
Step 3
x−4y+5z=0
(−2y+z)−4y+5z=0
−2y+z−4y+5z=0
−6y+6z=0
-y+z=0...(5)
therefore from equation (4) and (5), we can notice that the given system of equations has infinitely many solutions.
let z=k
therefore, from (4), we get
-y+z=0
-y+k=0
y=k
now substitute the value of y and z in x.
x=-2y+z
x=-2k+k
x=-k
Step 4
therefore, the values of x, y and z are -k, k and k respectively.
therefore all solutions of the given system of equations is x=-k, y=k and z=k

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