THIS USER ASKED 👇

Solve the system by elimination.. x+5y-4z=-10. 2x-y+5z=-9. 2x-10y-5z=0. a.) (5, –1, 0). b.) (–5, 1, 0). c.) (–5, –1, 0). d.) (–5, –1, –2).

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You have three equations assuming that the x, y, and z have the same value for all equations. The best way to solve this is to start by rearranging the first equation to isolate X.

X + 5y-4z = -10

X = 4z-5y-10

Then you do the same with the second equation to isolate ya.

2x-Y + 5z = -9

2x + 5z + 9 = Y.

Substitute the X equation for the second edited equation.

2 (4z-5y-10) + 5z + 9 = Y.

Simpligh

8z-10y-20 + 5z + 9 = Y.

13z-10y-11 = Y… 13z-11 = 11Y… Y = (13z-11) / 11

Substitute the equation X for the third equation.

2x-10y-5z = 0

2 (4z-5y-10) -10Y-5z = 0

8z-10y-20-10Y-5z = 0

Simpligh

3z-20y = 20

Enter the Y in the third equation

3z-20 (13z-11) / 11 = 20

3z- (260z + 220) / 11 = 20

Plural under 11 & Simplify

33z-260z + 220 = 220… -227z = 0

Divide this by setting Z = 0

Now put your value Z in the equation Y.

Y = (13 (0) -11) / 11

Y = 0-11 / 11

Y = -1

Finally, solve the first equation

X = 4z-5y-10

X = 4 (0) -5 (-1) -10

X = 0 + 5-10

X = -5

Your answer is choice C… (-5, -1, 0)