Answer:
x = -[tex]\frac{2}{3}[/tex], y = -[tex]\frac{1}{2}[/tex] and z = 1 is the solution of the system of the equations.
Step-by-step explanation:
The system of equation is given as
3x - 2y + z = 0 -----(1)
6x + 2y + 3z = -2 -------(2)
3x - 4y + 5z = 5 ------(3)
Now we add equations 1 from 2
(6x + 2y + 3z) + (3x - 2y + z) = 0 - 2
6x + 3x + 2y - 2y + 3z + z = -2
9x + 4z = -2 -----(4)
Further we multiply equation 1 by 2 and subtract from equation 3
(3x - 4y + 5z) - 2(3x - 2y + z) = 5 - 0
3x - 6x - 4y + 4y + 5z - 2z = 5
-3x + 3z = 5 -----(5)
Now we multiply equation 5 by 3 and add it to equation 4
3(-3x + 3z) + (9x + 4z) = 5×3 - 2
-9x + 9z + 9x + 4z = 15 -2
13z = 13
z = 1
Now we put z = 1 in equation number 5
-3x + 3×1 = 5
-3x + 3 = 5
-3x = 5 - 3
-3x = 2
x = -[tex]\frac{2}{3}[/tex]
Now we put the values of x and z in equation 1
3(-[tex]\frac{2}{3}[/tex] - 2y + 1 = 0
-2 - 2y + 1 = 0
-1 - 2y = 0
2y = -1
y = -[tex]\frac{1}{2}[/tex]
Therefore, x = -[tex]\frac{2}{3}[/tex], y = -[tex]\frac{1}{2}[/tex] and z = 1 is the solution of the system of the equations.
