[tex]a,\ b;\ c-the\ numbers\\\\\begin{array}{ccc}(1)\\(2)\\(3)\end{array} \left\{\begin{array}{ccc}a+b+c=7\\a-1=b+c\\a=4(b-c)\end{array}\right\\subtitute\ (3)\ to\ (2)\ and\ (1)\\ \left\{\begin{array}{ccc}4(b-c)+b+c=7\\4(b-c)-1=b+c\end{array}\right\\ \left\{\begin{array}{ccc}4b-4c+b+c=7\\4b-4c-1=b+c&|subtract\ from\ both\ sides\ "b"\ and\ "c"\&\ add \ 1\end{array}\right\\\left\{\begin{array}{ccc}5b-3c=7&|\multiply\ both\ sides\ by\ 3\\3b-5c=1&|both\ sides\ by\ (-5)\end{array}\right[/tex]
[tex]\underline{+\left\{\begin{array}{ccc}15b-9c=21\\-15b+25c=-5\end{array}\right}\\.\ \ \ \ \ 16c=16\ \ \ |divide\ both\ sides\ by\ 16\\.\ \ \ \ \boxed{c=1}\\\\Subtitute\ the\ value\ of\ c=1\ to\ 3b-5c=1\\\\3b-5\cdot1=1\\3b-5=1\ \ \ |add\ 5\ to\ both\ sides\\3b=6\ \ \ |divide\ both\ sides\ by\ 3\\\boxed{b=2}\\\\Subtitute\ the\ values\ of\ c=1\ and\ b=2\ to\ the\ a+b+c+=7\\\\a+2+1=7\\a+3=7\ \ \ |subtract\ 3\ from\ both\ sides\\\boxed{a=4}\\\\Answer:\boxed{4;\ 2;\ 1}[/tex]
