Verified Answer :

LammettHash

\(\begin{cases}x-6y-3z=4&(1)\\-2x-3z=-8&(2)\\-2x+2y-3z=-14&(3)\end{cases}\)

First eliminate \(y\) by adding \((1)\) to three times \((3)\). This gives

\((x-6y-3z)+3(-2x+2y-3z)=4+3(-14)\iff -5x-12z=-38\)

This reduces the system of one of two equations and two unknowns:

\(\begin{cases}-5x-12z=-38&(1)^*\\-2x-3z=-8&(2)^*\end{cases}\)

You can eliminate \(z\) by subtracting \((1)^*\) and four times \((2)^*\) to get

\((-5x-12z)-4(-2x-3z)=-38-4(-8)\iff3x=-6\implies x=-2\)

Back-substitute to find \(z\) and \(y\). You should end up with \((x,y,z)=(-2,-3,4)\).

Answer:

(-2,-3,4)

Step-by-step explanation: