Solving this system means solving each of the following equations:
\(\displaystyle{8}{x}+{6}={20},\)

\(\displaystyle{4}{y}−{12}=−{4},\)

and

\(\displaystyle{12}{z}+{18}={54}\)

You can obtain these equations by performing scalar multiplication on the vectors given. From here, you just use basic algebra to find that the solutions are as follows:

\(\displaystyle{x}={\frac{{{14}{\left\lbrace{8}\right\rbrace},{y}={2},{\quad\text{and}\quad}{z}={3}.}}{}}\)

\(\displaystyle{4}{y}−{12}=−{4},\)

and

\(\displaystyle{12}{z}+{18}={54}\)

You can obtain these equations by performing scalar multiplication on the vectors given. From here, you just use basic algebra to find that the solutions are as follows:

\(\displaystyle{x}={\frac{{{14}{\left\lbrace{8}\right\rbrace},{y}={2},{\quad\text{and}\quad}{z}={3}.}}{}}\)