[tex]\bigstar\boxed{\large\bf\red{\leadsto Age_{(father)}= 42\:yrs}}[/tex]
[tex]\bigstar\boxed{\large\bf\red{\leadsto Age_{(son)}=16\:yrs}}[/tex]
__________________________________________
[tex]\large\bf{\underline{According\:to\: situation\:one:}}[/tex]
➠ Let the age of son be x and the age of father be y
❒ Therefore :
[tex]\bf{⟼2x+10= y}[/tex] ⸻ (1)
[tex]\large\bf{\underline{According\:to\: situation\:two:}}[/tex]
[tex]\bf{⟼x^2 +y^2 = 4+3(x\times y)}[/tex]
➠ Substituting the value of y from equation (1)
[tex]\bf{⟹x^2 + (2x+10)^2 = 4+3(x\times ( 2x +10)}[/tex]
❒ Using identity :
[tex]\boxed{\arge\tt\pink{(a+b)^2 = a^2 + b^2 +2ab}}[/tex]
[tex]\bf{⟹x^2 + (2x)^2 +(10)^2 +2\times 2x\times 10 = 4+3(2x^2 +10x)}[/tex]
[tex]\bf{⟹x^2 + 4x^2 +100+40x = 4+6x^2 +30x }[/tex]
[tex]\bf{⟹5x^2 -6x^2 +40x -30x +100 -4 =0}[/tex]
[tex]\bf{⟹-x^2 +10x +96 =0}[/tex]
[tex]\large\bf{\underline{By\: splitting\: middle\:term:}}[/tex]
[tex]\bf{⟹-x^2 +16x -6x +96 }[/tex]
[tex]\bf{⟹-x(x-16)-6(x-96)}[/tex]
[tex]\bf{⟹(x-16)(-x-6)}[/tex]
➠ x- 16 =0
➜ x = 16
➠ -x-6 = 0
➜ x = -6
❒ Age cannot be negative, hence we take the value of x as 16
[tex]\large\bf{\underline{Therefore:}}[/tex]
➜ y = (2 × 16 )+ 10
➜ y = 32 + 10
➜ y = 42 yrs
