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).

You have three equations and assuming that the x, y, and z all 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 y.

2x-Y+5z=-9

2x+5z+9=Y

Substitute the X equation into the second rearranged equation.

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

Simplify

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

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

Substitute the X equation into the third equation.

2x-10y-5z=0

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

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

Simplify

3z-20y=20

Insert the Y into the third equation

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

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

Multiple by 11 & Simplify

33z-260z+220=220 … -227z=0

Divide this out and you will determine that Z =0 

Now put your Z value into the Y equation

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 option C … (-5, -1, 0)

step-by-step explanation:

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