system of equations,

Question

x+y+z= 6

2x-y+5z= 15

3x-y-2z= -5

Answer 1

Elimination method

x+y+z = 6 --------> (1)

2x-y+5z = 15 ----------> (2)

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

To eliminate the y value add the equations (1) and (2).

  x+y+z = 6

2x-y+5z = 15

___________

3x+6z = 21

Divide to each side by 3.

3x/3+6z/3 = 21/3

x+2z = 7---------> (4)

To eliminate the y value add the eequations (1) and (3)

  x+y+z = 6

3x-y-2z = -5

__________

4x-z = 1 ---------> (5)

Multiple to each side to the equation (5) by2.

8x-2z = 2 -------> (6)

To eliminate the z value add the equations (4) and (6).

 x+2z = 7

8x-2z = 2

_________

9x = 9

x = 9/9

⇒ x = 1

Substitute the x value in (4).

1+2z = 7

Subtract 1 from each side.

1+2z-1 = 7-1

2z = 6

Divide to each side by 2.

2z/2 = 6/2

⇒ z = 3

Substitute the x ,z in equation (1).

1+y+3 = 6

4+y = 6

Subtract 4 from each side.

4+y-4 = 6-4

⇒ y = 2

Solution of given system is x = 1, y = 2, z = 3.