Solving $\frac{5}{1+i_1} + \frac{105}{(1+i_2)^2} = 100,96$ and $\frac{6}{1+i_1} + \frac{106}{(1+i_2)^2} = 102,84$

Question

I didn't have linear equations for a long time and I struggle to solve this one:

$$A = \frac{5}{1+i_1} + \frac{105}{(1+i_2)^2} = 100,96$$

$$B = \frac{6}{1+i_1} + \frac{106}{(1+i_2)^2} = 102,84$$

Apparently $1+i_1 \approx 1,036$ and $1+i_2 \approx 1,045$, but I struggle to get there.

I tried expanding the fraction to

$$\frac{5(1+i_2)^2}{(1+i_1) (1+i_2)^2} + \frac{105 (1+i_1)}{(1+i_2)^2 (1+i_1)} = 100,96$$, but couldn't really get anywhere.

Can someone tell me what the approach is here? Thanks in advance!

Answer 1

As lulu says, define $x=\frac 1{1+i_1}, y=\frac 1{(1+i_2)^2}$ to save writing. We have $$5x+105y=100.96\\ 6x+106y=102.84\\x+y=2.88$$ so substitute $y=2.88-x$ into one of them and you have a linear equation. Once you have $x,y$ you can get $i_1,i_2$