Question

Select the correct answer. Which system of equations has a solution of (-2,-2,-2) ​

Answer 1

[B]
`\begin{bmatrix}x+2y=-6\\ y+2z=-6\\ x-y-z=2\end{bmatrix}`

Explanation:

Going through all answer choice to find the solution:

[A]
`\begin{bmatrix}x+y=0\\ y-z=-2\\ x+y-z=-4\end{bmatrix}`

`\mathrm{Isolate\;x\;for\;x+y=0;x=-y}`

`\mathrm{Substitute\:}x=-y`

`\begin{bmatrix}y-z=-2\\ -y+y-z=-4\end{bmatrix}`

`\mathrm{Simplify}`

`\begin{bmatrix}y-z=-2\\ -z=-4\end{bmatrix}`

`\mathrm{Isolate\;z\;for\;-z=-4;z=4}`

`\mathrm{Substitute\:}z=4`

`\begin{bmatrix}y-4=-2\end{bmatrix}`

`\mathrm{Isolate\;y\;for\;y-4=-2;y=2}`

`\mathrm{For\:}x=-y`

`\mathrm{Substitute\:}z=4,\:y=2`

`x=-2`

`\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}`

`x=-2,\:z=4,\:y=2`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

[B]
`\begin{bmatrix}x+2y=-6\\ y+2z=-6\\ x-y-z=2\end{bmatrix}`

`\begin{bmatrix}y+2z=-6\\ z-y-z=2\end{bmatrix}`

`\mathrm{Substitute\:}y=-2`

`\begin{bmatrix}-2+2z=-6\end{bmatrix}`

`\mathrm{For\:}x=-6-2y`

`\mathrm{Substitute\:}z=-2,\:y=-2`

`x=-6-2\left(-2\right)`

`x=-2`

`\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}`

`x=-2,\:z=-2,\:y=-2`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

[C]
`\begin{bmatrix}3x-y=-8\\ y-3z=-8\\ x+y+z=-8\end{bmatrix}`

`\mathrm{Substitute\:}x=(-8+y)/(3)`

`\begin{bmatrix}y-3z=-8\\ (-8+y)/(3)+y+z=-8\end{bmatrix}`

`\mathrm{Simplify}`

`\begin{bmatrix}y-3z=-8\\ z+(-8+4y)/(3)=-8\end{bmatrix}`

`\mathrm{Substitute\:}y=-8+3z`

`\begin{bmatrix}z+(-8+4\left(-8+3z\right))/(3)=-8\end{bmatrix}`

`\mathrm{Simplify}`

`\begin{bmatrix}(15z-40)/(3)=-8\end{bmatrix}`

`\mathrm{For\:}y=-8+3z`

`\mathrm{Substitute\:}z=(16)/(15)`

`y=-8+3\cdot (16)/(15)`

`y=-(24)/(5)`

`\mathrm{For\:}x=(-8+y)/(3)`

`\mathrm{Substitute\:}z=(16)/(15),\:y=-(24)/(5)`

`x=(-8-(24)/(5))/(3)=x=-(64)/(15)`

`\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}`

`x=-(64)/(15),\:z=(16)/(15),\:y=-(24)/(5)`

`x=-(64)/(15)=-4.26666666667, z=(16)/(15)=1.06666666667,y=-(24)/(5)=-4.8`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

[D]
`\begin{bmatrix}x-2y+z=0\\ 2y+9z=-20\\ x-y+z=0\end{bmatrix}`

`\mathrm{Substitute\:}y=(-20-9z)/(2)`

`\begin{bmatrix}x-2\cdot (-20-9z)/(2)+z=0\\ x-(-20-9z)/(2)+z=0\end{bmatrix}`

`\mathrm{Simplify}`

`\begin{bmatrix}x+20+10z=0\\ x+(20+11z)/(2)=0\end{bmatrix}`

`\mathrm{Substitute\:}x=-10z-20`

`\begin{bmatrix}-10z-20+(20+11z)/(2)=0\end{bmatrix}`

`\mathrm{Simplify}`

`\begin{bmatrix}(20-9z)/(2)-20=0\end{bmatrix}`

`\mathrm{For\:}x=-10z-20`

`\mathrm{Substitute\:}z=-(20)/(9)`

`x=-10\left(-(20)/(9)\right)-20`

`x=(20)/(9)`

`\mathrm{For\:}y=(-20-9z)/(2)`

`\mathrm{Substitute\:}x=(20)/(9),\:z=-(20)/(9)`

`y=(-20-9\left(-(20)/(9)\right))/(2)=0`

`y=0`

`\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}`

`x=(20)/(9),\:z=-(20)/(9),\:y=0`

`x=(20)/(9)=2.22222222222,\:z=-(20)/(9)-2.22222222222,\:y=0`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Kavinsky