What is the answer to 8x - 6z = 38 , 2x - 5y + 3z = 5 , x + 10y - 4z = 8?

Question

Agebra 2; system of equations using three variables.

Answer 1

Elmination method

Given first equation 8x+0*y-6z  = 38 --------> (1)

And second equaton 2x-5y+3z = 5 ----------> (2)

Muliple to each side by negitive 4.

 -8x+20y-12z = -20 ---------> (3)

And third equation x+10y-4z = 8

Muliple to each side by negitive 2.

-2x-20y+8z = -16 ---------> (4)

To elminate the x value, add the equations (1) and (3).

8x+0*y-6z  = 38

-8x+20y-12z = -20

_______________

20y-18z = 18 ------------> (5)

Muliple to each side by 25.

500y-450z = 450 ---------> (6)

To elminate the x value, add the equations (2) and (4).

  2x-5y+3z = 5

-2x-20y+8z = -16

______________

-25y+11z = -11

Muliple to each side by 20.

-500y+220z = -220 -----------> (7)

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

500y-450z = 450

-500y+220z = -220

________________

-230z = 230

Dvide to each side by negitive 230.

-230z/-230 = 230/-230

⇒ z = -1

Substitute the z value in equation (1).

8x-6*-1 = 38

8x+6 = 38

Subtract 6 from each side.

8x+6-6 = 38-6

8x = 32

Dvide to each side by 8.

8x/8 = 32/8

⇒ x = 4

Substitute the x, z  value in equation (2).

2*4-5y+3*-1 = 5

8-5y-3 = 5

-5y+5 = 5

Subtract 5 from each side.

-5y+5-5 = 5-5

-5y = 0

Dvide to each side by negitive 5.

-5y/-5 = 0/-5

⇒ y = 0

Solution x = 4, y = 0, z = -1.