2x-6y+6z=17, x+2y-3z=13 and 3x+21z=26

Question

am trying to find what the x , y , and z value of the equation would be.

Answer 1

Elimination method

Given equations are 2x-6y+6z = 17 ---> (1)

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

3x+21z = 26 ---> (3)

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

3x+6y-9z = 39 ---> (4)

To eliminate the y value add the equations (1)&(4).

2x-6y+6z = 17

3x+6y-9z = 39

____________

5x-3z = 56 ---> (5)

Multiple to each side of equation (5) by 7.

35x-21z = 392 ---> (6)

To eliminate the z value add the equations (6)&(3).

35x-21z = 392

3x+21z = 26

____________

38x = 418

Divide to each side by 38.

38x/38 = 418/38

x = 11

Substitute the x value in (3).

33+21z = 26

Subtract 33 from each side.

33-33+21z = 26-33

21z = -7

Divide to each side by 21.

21z/21 = -7/21

z = -1/3

Substitute x,z in (2).

11+2y+1 = 13

2y+12 = 13

Subtract 12 from each side.

2y+12-12 = 13-12

2y = 1

Divide to each side by 2.

2y/2 = 1/2

y = 1/2.

Solution x = 11, y = 1/2, z = -1/3.