Question

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

Answer 1

Answer:

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

Step-by-step 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] $\begin{bmatrix}3x-y=-8\\ y-3z=-8\\ x+y+z=-8\end{bmatrix}$

$\mathrm{Substitute\:}x=\frac{-8+y}{3}$

$\begin{bmatrix}y-3z=-8\\ \frac{-8+y}{3}+y+z=-8\end{bmatrix}$

$\mathrm{Simplify}$

$\begin{bmatrix}y-3z=-8\\ z+\frac{-8+4y}{3}=-8\end{bmatrix}$

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

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

$\mathrm{Simplify}$

$\begin{bmatrix}\frac{15z-40}{3}=-8\end{bmatrix}$

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

$\mathrm{Substitute\:}z=\frac{16}{15}$

$y=-8+3\cdot \frac{16}{15}$

$y=-\frac{24}{5}$

$\mathrm{For\:}x=\frac{-8+y}{3}$

$\mathrm{Substitute\:}z=\frac{16}{15},\:y=-\frac{24}{5}$

$x=\frac{-8-\frac{24}{5}}{3}=x=-\frac{64}{15}$

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

$x=-\frac{64}{15},\:z=\frac{16}{15},\:y=-\frac{24}{5}$

$x=-\frac{64}{15}=-4.26666666667, z=\frac{16}{15}=1.06666666667,y=-\frac{24}{5}=-4.8$

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

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

$\mathrm{Substitute\:}y=\frac{-20-9z}{2}$

$\begin{bmatrix}x-2\cdot \frac{-20-9z}{2}+z=0\\ x-\frac{-20-9z}{2}+z=0\end{bmatrix}$

$\mathrm{Simplify}$

$\begin{bmatrix}x+20+10z=0\\ x+\frac{20+11z}{2}=0\end{bmatrix}$

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

$\begin{bmatrix}-10z-20+\frac{20+11z}{2}=0\end{bmatrix}$

$\mathrm{Simplify}$

$\begin{bmatrix}\frac{20-9z}{2}-20=0\end{bmatrix}$

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

$\mathrm{Substitute\:}z=-\frac{20}{9}$

$x=-10\left(-\frac{20}{9}\right)-20$

$x=\frac{20}{9}$

$\mathrm{For\:}y=\frac{-20-9z}{2}$

$\mathrm{Substitute\:}x=\frac{20}{9},\:z=-\frac{20}{9}$

$y=\frac{-20-9\left(-\frac{20}{9}\right)}{2}=0$

$y=0$

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

$x=\frac{20}{9},\:z=-\frac{20}{9},\:y=0$

$x=\frac{20}{9}=2.22222222222,\:z=-\frac{20}{9}-2.22222222222,\:y=0$

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

Kavinsky