Question

Solve the following system.

Answer 1

Answer:

$y=-2,\:\\x=1,\:\\z=-3$

Step-by-step explanation:

$\begin{bmatrix}y+4z=-14\\ x+3z=-8\\ 2x-3y=8\end{bmatrix}\\\\Isolate\:y\:for\:y+4z =-14\: : \:y=-14-4z\\\\\mathrm{Subsititute\:}y=-14-4z\\\\\begin{bmatrix}x+3z=-8\\ 2x-3\left(-14-4z\right)=8\end{bmatrix}\\\\Isolate\:x\:for\:x+3z =-8\: :\:x =-8-3z\\\mathrm{Subsititute\:}x=-8-3z\\\\\begin{bmatrix}2\left(-8-3z\right)-3\left(-14-4z\right)=8\end{bmatrix}\\\\Simplify\\\\\begin{bmatrix}6z+26=8\end{bmatrix}\\\\Isolate\:z\:for 6z+26=8 \: ;\:z =-3\\\\\mathrm{For\:}x=-8-3z$

$\mathrm{Subsititute\:}z=-3\\\\x=-8-3\left(-3\right)\\\\-8-3\left(-3\right)=1\\\\x=1\\\\\mathrm{For\:}y=-14-4z\\\\\mathrm{Subsititute\:}x=1,\:z=-3\\\\y=-14-4\left(-3\right)\\\\-14-4\left(-3\right)=-2\\\\y=-2\\\\\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}\\\\y=-2,\:x=1,\:z=-3$