Question

Two years ago, father was three times as old as his son and two years hence, twice his age will be equal to five times that of his son. Find their present ages ​

Answer 1

$\sf Let\: the \:age\; of \: son \: be \:x\:.\\\sf Two \:years \:ago \:,\:son's\: age= x-2.\\\sf And \: Father's Age = 3(x-2)\\\sf The \:present \;age\: of\: Father = 3x-6+2=3x-4\\\sf Two\: years\: hence\:,\: father's\: age=3x-4+2=3x-2\\\sf And \: Son's \: age = x+2$

$\sf \Longrightarrow 5 \times (x + 2) = 2 \times (3x - 2)\\\\\sf \Longrightarrow 5x + 10 = 6x - 4\\\\\sf \Longrightarrow 6x-5x = 10+4\\\\\sf \Longrightarrow x = 14$

$\tt \bf Present \:of\: Son = 14 \:years\\\\\tt \bf Present \: age \: of \: Father = 38\:years$

HOPE IT HELPS U .............

Answer 2

Let the ages of the father and son be x and y respectively.

Given

Two years ago, father was three times as old as his son.

$x-2=3(y-2)\\\implies x-2=3y-6\\\implies x-3y=-6+2\\\implies x-3y=-4--(1)$

Two years hence, twice his age will be equal to five times that of his son.

$2(x+2)=5(y+2)\\\implies 2x+4=5y+10\\\implies 2x-5y=10-4\\\implies 2x-5y=6--(2)$

Multiplying the first equation by 1, we get,

$\implies 2x-6y=-8--(3)$

Solving (2) and (3), we get,

$2x-5y=6\\\underline{2x-6y=-8}\\\underline{\underline{y=14}}$

Substituting y in the first equation, we get,

$x-42=-4\\\implies x = -4+42\\\implies x = 38$

$\boxed{\boxed{\bold{Therefore, \ the \ father's \ and \ the \ son's \ age \ is \ 38 \ years \ and \ 14 \ years \ respectively}}}}}}}}}$