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2x-6y+9z=-8
5x+y+2z=10
3x+y-8z = -28

1 Answer

1 vote

Explanation:

I assume we have to find the solution of this system of equations.

2x-6y+9z=-8

5x+y+2z=10

3x+y-8z = -28

let's subtract the third from the second equation to get rid of y :

5x + y + 2z = 10

- 3x + y - 8z = -28

-----------------------

2x + 0 + 10z = 38

2x + 10z = 38

x + 5z = 19

x = 19 - 5z

next, let's multiply the 3rd equation by 6 and add it to the first equation :

2x - 6y + 9z = - 8

18x + 6y - 48z = -168

---------------------------------

20x 0 - 39z = -176

now, we use

x = 19 - 5z in

20x - 39z = -176

20(19 - 5z) - 39z = -176

380 - 100z - 39z = -176

556 - 139z = 0

556 = 139z

z = 556/139 = 4

x = 19 - 5z = 19 - 5×4 = 19 - 20 = -1

now we use one of the original squadrons to get y.

e.g.

5×-1 + y + 2×4 = 10

-5 + y + 8 = 10

3 + y = 10

y = 7

x = -1

y = 7

z = 4

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