We have the following system:
[tex]\begin{gathered} x=6y-2\ldots(A) \\ 2x+5y=47\ldots(B) \end{gathered}[/tex]Solving by substitution method.
If we substitute equation A into B, we get
[tex]2(6y-2)+5y=47[/tex]which gives
[tex]12y-4+5y=47[/tex]By combining similar terms, we have
[tex]\begin{gathered} 12y+5y-4=47 \\ 17y-4=47 \end{gathered}[/tex]If we move -4 to the right hand side, we obtain
[tex]\begin{gathered} 17y=47+4 \\ 17y=51 \end{gathered}[/tex]Then, if we move the coefficient of y to the right hand side, wehave
[tex]\begin{gathered} y=\frac{51}{17} \\ y=3 \end{gathered}[/tex]Then, we can substitute this result into equation A, which gives
[tex]x=6(3)-2[/tex]so, xi given by
[tex]\begin{gathered} x=18-2 \\ x=16 \end{gathered}[/tex]Therefore, the answer is
[tex]\begin{gathered} x=16 \\ y=3 \end{gathered}[/tex]