Question

Find the values of x,y, and z that satisfies the system: (2x-5y+2z=-18, -2x+3y-3z= 14, x-y+5z= -6)

Answer 1

2x-5y+2z = -18 -------> (1)

-2x+3y-3z = 14 ----------> (2)

x-y+5z = -6 ---------> (3)

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

2x-5y+2z = -18

-2x+3y-3z = 14

____________

-2y-z = -4 ----------> (4)

Multiple to each side the equation (3) by 2.

2x-2y+10z  = -12 --------> (5)

To eliminate the x value subtracrt equation (5) from (1).

2x-2y+10z = -12

2x-5y+2z = -18

(-)  (+) (-)   (+)

____________

3y+8z  = 6 -----------> (6)

Multiple to each side the equation (4) by 8.

-16y-8z = -32 ----------> (7)

To eliminate the y value add the equations (6),(7).

   3y+8z = 6

-16y-8z = -32

____________

-13y = -26

Divide to each side by negitive 13.

-13y/-13 = -26/-13

y = 2

Substitute the y value in (6).

3(2)+8z = 6

6+8z = 6

Subtrcr 6 from each side.

6-6+8z = 6-6

8z = 0

z = 0

Substitute the y,z values in (3).

x-2+0 = -6

x-2 = -6

Add 2 to each side.

x-2+2 = -6+2

x = -4

Solution of given system is x = -4,y = 2, z = 0.