\n– Let the present age of the son be $S$ years and the present age of the father be $F$ years.
\n– One year ago, the son’s age was $S-1$ and the father’s age was $F-1$.
\n– According to the first condition: $F-1 = 4(S-1)$. This simplifies to $F-1 = 4S – 4$, so $F = 4S – 3$ (Equation 1).
\n– After six years, the son’s age will be $S+6$ and the father’s age will be $F+6$.
\n– According to the second condition: The father’s age ($F+6$) exceeds twice his son’s age ($2(S+6)$) by 9 years. So, $F+6 = 2(S+6) + 9$.
\n– This simplifies to $F+6 = 2S + 12 + 9$, which is $F+6 = 2S + 21$. So, $F = 2S + 15$ (Equation 2).
\n– Now we have a system of two linear equations for $F$ and $S$:
\n 1) $F = 4S – 3$
\n 2) $F = 2S + 15$
\n– Equating the expressions for $F$: $4S – 3 = 2S + 15$.
\n– Subtract $2S$ from both sides: $2S – 3 = 15$.
\n– Add 3 to both sides: $2S = 18$.
\n– Divide by 2: $S = 9$. The son’s present age is 9 years.
\n– Substitute $S=9$ into Equation 1 (or Equation 2) to find $F$:
\n $F = 4(9) – 3 = 36 – 3 = 33$. The father’s present age is 33 years.
\n– The ratio of their present age (Father : Son) is $F : S = 33 : 9$.
\n– This ratio can be simplified by dividing both numbers by their greatest common divisor, which is 3.
\n $33 \\div 3 = 11$
\n $9 \\div 3 = 3$
\n– The simplified ratio is 11 : 3.
\n<\/section>\n