Question

Solve the system of elimination -2x+2y+3z=0 -2x-y+z=-3 2x+3y+3z=5

Answer 1

x = 1 , y = 1 and z = 0

Explanation:

-2x+2y+3z=0 ------(1)

-2x-y+z=-3 --------(2)

2x+3y+3z=5 ---------(3)

To find the values of x,y and z

(3) - (2)⇒ 2y + 4z = 2

y + 2z = 1 --------(4)

(1) - (2)⇒ 3y +2z =3 ---(5)

(5) - (4)⇒ 2y = 2

y = 1

Substituting the value of y in (4) we get z =0

Substituting the value of y and z in (1) we get x = 1

Therefore x = 1 , y = 1 and z = 0