4x+3y+z=-26, x-3y+2z=21, 11x-2y+3z=-19 by elimination

Question

Need to solve for x, y ,z using elimination process of equations

Answer 1

Elimination method

4x+3y+z = -26 -----------> (1)

x-3y+2z = 21  -------------> (2)

11x-2y+3z = -19 -----------> (3)

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

4x+3y+z = -26

x-3y+2z = 21

___________

5x+3z = -5 ---------> (4)

Multiply to each side equation (2) by2 and equation (3) by 3.

2x-6y+4z = 42 ------------> (5)

33x-6y+9z = -57-----------> (6)

To eliminate y subtract equation (6) from (5)

2x-6y+4z = 42

33x-6y+9z = -57

(-)  (+)  (-)  = (+)

_____________

-31x -5z = 99 ----------> (7)

Multiply to each side equation (4) by 5 and equation (7) by 3.

And to eliminate z add the equations.

25x+15z = -25

-93x-15z = 297

____________

-68x = 272

Divide to each side by negitive 68.

-68x/-68 = 272/-68

⇒ x = -4

Substitute the value of x  in equation (4).

5*-4+3z = -5

-20+3z = -5

Add 20 to each side.

-20+20+3z = -5+20

3z = 15

Divide to each side by 3.

3z/3 = 15/3

⇒ z = 5

Substitute thevalues of x,z in equation (2)

-4-3y+2*5 = 21

-4-3y+10 = 21

-3y+6 = 21

Subtract 6 from each side.

-3y+6-6 = 21-6

-3y = 15

Divide to each side by negitive 3.

-3y/-3 = 15/-3

⇒ y = -5

Solution of given system is x =-4, y = -5, z = 5.