We have:
\(
y=3x+2~~(i)\\\\
-4x+2y=8~~(ii)
\)
Substituting (i) in (ii):
\(
-4x+2y=8\\\\
-4x+2(3x+2)=8\\\\
-4x+6x+4=8\\\\
2x+4=8\\\\
2x=4\\\\
x=\dfrac{4}{2}\\\\
\boxed{x=2}
\)
Using the value obtained above in the expression (i):
\(
y=3x+2\\\\
y=3\cdot2+2\\\\
y=6+2\\\\
\boxed{y=8}
\)
