The solution of -2x + 2y + 3z = 0, -2x - y + z = -3, 2x + 3y + 3z = 5 is x = 1 , y = 1 and z = 0.
Solution:
Need to solve given system of equation using elimination method.
-2x + 2y + 3z = 0 ------(1)
-2x - y + z = -3 ------(2)
2x + 3y + 3z = 5 ------(3)
As coefficients of x is same in (1) and (2) , subtracting (1) from (2) to eliminate variable x
(-2x - y + z) - (-2x + 2y +3z ) = (-3) - 0
=> -y + z – 2y – 3z = -3
=> -3y – 2z = -3
=> 3y + 2z = 3 --------(4)
On Adding equation 2 and 3 , we get our other equation in two variables y and z.
-2x - y + z + 2x + 3y + 3z = -3 + 5
=> 2y + 4z = 2
=> y + 2z= 1 ---------(5)
As coefficients of z is same in (4) and (5), subtracting (4) from (5) to eliminate variable z
(y + 2z) – (3y + 2z) = 1 – 3
=> -2y = -2
=> y = 1
Substituting y = 1 in equation 5, to get value of z
=> 1 + 2z = 1
=> 2z = 1 – 1 = 0
=> z = 0
Substituting y = 1 and z = 0 in equation (1) to get value of x
=> -2x + (2 x 1) + 3 x 0 = 0
=>-2x +2 = 0
=> -2x = -2
=> x = 1
Hence solution of given system is x = 1 , y = 1 and z = 0.