Let x be the number of nickels that Nicholas has, y the number of dimes, and z the number quarters. Using the given information we can set the following system of equations:
[tex]\begin{gathered} 0.05x+0.10y+0.25z=12.30, \\ y=3+2x, \\ z=3x\text{.} \end{gathered}[/tex]Substituting the second and third equations in the first one we get:
[tex]0.05x+0.10(3+2x)+0.25(3x)=12.30.[/tex]Solving for x we get:
[tex]\begin{gathered} 0.05x+0.30+0.20x+0.75x=12.30, \\ 1x=12.30-0.30, \\ x=12. \end{gathered}[/tex]Substituting x=12 in the second and third equations on the board, we get:
[tex]\begin{gathered} y=3+2(12)=3+24=27. \\ z=3(12)=36. \end{gathered}[/tex]Answer: Nicholas has 12 nickels, 27 dimes, and 36 quarters.