Answer:[tex]\dfrac{110}{13},\dfrac{290}{13}[/tex]
Step-by-step explanation:
Let [tex]a_{I}[/tex] be the present age of I.
Let [tex]a_{p}[/tex] be the present age of P.
[tex]5[/tex] years ago P was [tex]5[/tex] times older than l.
Age of P [tex]5[/tex] years ago is [tex]a_{p}-5[/tex]
Age of I [tex]5[/tex] years ago is [tex]a_{I}-5[/tex]
[tex]5\times (a_{I}-5)=a_{p}-5\\5a_{I}=a_{p}+20[/tex] ....(i)
After [tex]10[/tex] years l will be [tex]\frac{4}{7}[/tex] as old as P.
Age of P [tex]10[/tex] years after is [tex]a_{p}+10[/tex]
Age of I [tex]10[/tex] years after is [tex]a_{I}+10[/tex]
[tex]a_{I}+10=\frac{4}{7}\times (a_{p}+10)\\ 7a_{I}+30=4a_{p}[/tex] ...(ii)
using (i) and (ii),
[tex]7a_{I}+30=4(5a_{I}-20)\\110=13a_{I}\\a_{I}=\frac{110}{13}years[/tex]
[tex]a_{P}=5a_{I}-20=5\times \frac{100}{13}-20=\frac{290}{13} years[/tex]
