x-y+z=-1, 5x+2y+6z=3, -2=3y+4z=1

Question

Solve using any method for the values of x, y and z

Answer 1

Elimination method :

Consider the system of equations are

x - y + z = - 1          → ( 1 )

5x + 2y + 6z = 3      → ( 2 )

- 2x + 3y + 4z = 1    → ( 3 )

Write the equations ( 1 ) & ( 2 ) in column form, and Multiply equation (1 )  by 2, then add them, to eliminate the y variable.

2x - 2y + 2z = - 2

5x + 2y + 6z = 3

(+)____________

7x + 8z = 1          → ( 4 )

Write the equations ( 2 ) & ( 3 ) in column form, and Multiply eq (2 )  by 3, and eq ( 3 ) by 2, then subtact them, to eliminate the y variable.

15x + 6y + 18z = 9

- 4x + 6y + 8z = 2

( - )_____________

19x + 10z = 7       → ( 5 )

Write the equations ( 4 ) & ( 5 ) in column form, and Multiply eq (4 )  by 10, and eq ( 5 ) by 8, then subtact them, to eliminate the z variable.

70x + 80z = 10

152x + 80z = 56

( - )___________

- 82x = - 46

⇒ x = 46/82 = 23/41.

Substitute the value x = 23/41 in eq ( 4 ) : 7x + 8z = 1 and solve for z.

7(23/41) + 8z = 1

⇒ z = - 15/41.

Substitute the values x = 23/41 and z = - 15/41 in eq ( 1 ) : x - y + z = - 1, and solve for y.

(23/41) - y + (- 15/41) = - 1

y = (23/41) - (15/41) + 1 = (23 - 15 + 41)/41

⇒ y = 49/41.

The solution of the system  is x = 23/41, y = 49/41, and z = - 15/41.