Answer:
The given system of equations is :
2x + 3y + 5z = 1 -> eq. (1)
3x + 2y + z = 5 -> eq. (2)
Multiplying eq. (1) with 3 and eq. (2) with 2, we get :
6x + 9y + 15z = 3 -> eq. (3)
and, 6x + 4y + 2z = 10 -> eq. (4)
respectively.
Subtracting eq. (4) from eq. (3) :
5y + 13z = -7 -> eq. (5)
Multiplying eq. (1) with 2 and eq. (2) with 3, we get :
4x + 6y + 10z = 2 -> eq. (6)
and, 9x + 6y + 3z = 15 -> eq. (7)
respectively.
Subtracting eq. (7) from eq. (6) :
-5x + 7z = -13 -> eq. (8)
Let z = t, where t is a parameter.
We can write eq. (5) and eq. (8) as :
5y + 13t = -7 & -5x + 7t = -13
i.e., 5y = -7 - 13t & 5x = 13 + 7t
y = -(7/5) - (13t/5) & x = (13/5) + (7t/5)
Therefore, the solution for the given system of equations is :
x = (13/5) + (7t/5)
y = -(7/5) - (13t/5)
z = t
where, t can take any value.
Explanation: