Answer:
the values of x, y and z are x= 2, y =-1 and z=1
Explanation:
We need to solve the following system of equations.
We will use elimination method to solve these equations and find the values of x, y and z.
2x + 2y + 5z = 7 eq(1)
6x + 8y + 5z = 9 eq(2)
2x + 3y + 5z = 6 eq(3)
Subtracting eq(1) and eq(3)
2x + 2y + 5z = 7
2x + 3y + 5z = 6
- - - -
_____________
0 -y + 0 = 1
-y = 1
=> y = -1
Subtracting eq(2) and eq(3)
6x + 8y + 5z = 9
2x + 3y + 5z = 6
- - - -
______________
4x + 5y +0z = 3
4x + 5y = 3 eq(4)
Putting value of y = -1 in equation 4
4x + 5y = 3
4x + 5(-1) = 3
4x -5 = 3
4x = 3+5
4x = 8
x= 8/4
x = 2
Putting value of x=2 and y=-1 in eq(1)
2x + 2y + 5z = 7
2(2) + 2(-1) + 5z = 7
4 -2 + 5z = 7
2 + 5z = 7
5z = 7 -2
5z = 5
z = 5/5
z = 1
So, the values of x, y and z are x= 2, y =-1 and z=1