Question

Solve for the value of x. 2x-y+z=6, x-3y-2z=13  and  2x-3y-3z=16.

 

Answer 1

Elimination method

Given equations are 2x-y+z = 6 ---> (1)

x-3y-2z = 13 ---> (2)

2x-3y-3z = 16 ---> (3)

To eliminate the y value subtract the equation (2) from (3).

2x-3y-3z = 16

x-3y-2z = 13

(-)(+) (+)  (-)

____________

x-z = 3 ---> (4)

Mulitiple to each side of equation (1) by negitive 3.

-6x+3y-3z = -18 ---> (5)

To eliminate the y value add the equations (3)&(5).

2x-3y-3z = 16

-6x+3y-3z = -18

_______________

-4x-6z = -2 ---> (6)

Mulitiple to each side of equation (4) by 4.

4x-4z = 12 ---> (7)

To eliminate the x value add the equatons (6)&(7).

-4x-6z = -2

4x-4z = 12

__________

-10z = 10

Divide to each side by negitive 10.

-10z/-10 = 10/-10

z = -1

Substitute the z value in (2).

x-3y+2 = 13

x-3y = 11 ---> (8)

Substitute the z value in (3).

2x-3y+3 = 16

2x-3y = 13 ---> (9)

To eliminate y value subtract (9) from (8).

x-3y = 11

2x-3y = 13

(-)(+)    (-)

__________

-x = -2

x = 2

Substitute x,z value in (1).

4-y-1 = 6

-y+3 = 6

-y = 6-3

-y = 3

y = -3

Solution x = 2.