Question 754747
P=1400
I=80
Assume t=1

{{{I[1]+I[2]=80}}}

or...


{{{0.06P[1]+0.04P[2]=80}}}

for convenience we multiply by 50


{{{3P[1]+2P[2]=4000}}}


and the total principal is:



{{{P[1]+P[2]=1400}}}

then we double each term


{{{2P[1]+2P[2]=2800}}}




Next we subtract to solve


{{{3P[1]+2P[2]=4000}}}
{{{2P[1]+2P[2]=2800}}}


{{{P[1]=1200}}}


and thus

{{{P[2]=200}}}




:)