Question

The sum of three numbers is 7. One of the numbers is one more than the sum of the other two numbers. It is also four times the difference between the two numbers.

Answer 1

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

Answer 2

The only numbers I could think of would be 1, 2, and 4 because 4 is greater than 3 (2+1) and four times greater than 1 (2-1)