Let x, and y be the two numbers the problem is asking for, then we can set the following system of equations:
[tex]\begin{gathered} x+y=30, \\ \frac{x}{y}=2. \end{gathered}[/tex]Solving the second equation for x, we get:
[tex]\begin{gathered} \frac{x}{y}\cdot y=2\cdot y, \\ x=2y\text{.} \end{gathered}[/tex]Now, substituting x=2y in the first equation we get:
[tex]2y+y=30.[/tex]Solving for y we get:
[tex]\begin{gathered} 3y=30, \\ \frac{3y}{3}=\frac{30}{3}, \\ y=10. \end{gathered}[/tex]Finally, we substitute y=10 in the first equation on the board and get:
[tex]x+10=30.[/tex]Solving for x we get:
[tex]\begin{gathered} x=30-10, \\ x=20. \end{gathered}[/tex]Answer: the numbers we are looking for are 20 and 10.