what is the solution of the system? use elimination:(x+y-3Z=8),(-x-2y+z=4) and (-x+y+4z=7)

Question

how can i work this problem out?

Answer 1

Elimination method

Given equations are x+y-3z = 8 ---> (1)

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

-x+y+4z = 7---> (3)

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

x+y-3z = 8

-x-2y+z = 4

___________

-y-2z = 12 ---> (4)

To eliminate the x value add the equations (1)&(3).

x+y-3z = 8

-x+y+4z = 7

___________

2y+z = 15

Multiple to each side by 2.

4y+2z = 30 ---> (5)

To eliminatethe z value add the equations (5)&(4).

4y+2z = 30

-y-2z = 12

___________

3y = 42

Divide to each side by 3.

3y/3 = 42/3

⇒ y = 14

Substitute the y value in (4).

-14-2z = 12

Add 14 to each side.

-14+14-2z = 12+14

-2z = 26

Divide to each side by negitive 2.

-2z/-2 = 26/-2

⇒ z = -13

Substitute the y,z values in (1).

x+14+39 = 8

x+53 = 8

Subtract 53 from each side.

⇒ x = -45

Solution x = -45, y = 14, z = -13.