Answer :
[tex]Other\ method:\\\\x;y-searches\\\\ \left\{\begin{array}{ccc}x+y=30&|subtract\ y\ from\ both\ sides\\x:y=3:2\end{array}\right\\\left\{\begin{array}{ccc}x=30-y&(1)\\\frac{x}{y}=\frac{3}{2}&(2)\end{array}\right\\\\subtitute\ (1)\ to\ (2):\\\\\frac{30-y}{y}=\frac{3}{2}\ \ \ \ |cross\ multiply\\\\(3)(y)=(2)(30-y)\\3y=60-2y\ \ \ \ |add\ 2y\ to\ both\ sides\\5y=60\ \ \ \ |divide\ both\ sides\ by\ 5\\\boxed{y=12}[/tex]
[tex]Put\ the\ value\ of\ y\ to\ (1):\\x=30-12\\\boxed{x=18}\\\\Answer:\boxed{\left\{\begin{array}{ccc}x=18\\y=12\end{array}\right\to18:12}[/tex]
[tex]Put\ the\ value\ of\ y\ to\ (1):\\x=30-12\\\boxed{x=18}\\\\Answer:\boxed{\left\{\begin{array}{ccc}x=18\\y=12\end{array}\right\to18:12}[/tex]
