$100 + [110/(1+r)] = [1/ (1+r)] + [(232 /(1+r)^2 ]$

Question

Need to learn how to solve this: $100 + \frac{110}{1 + r} = \frac{1}{1 + r} + \frac{232}{(1 + r)^{2}}$. Checked this site got to the 3rd line and am completely lost. Can someone help me solve for r line by line please?

Answer 1

You have this:

$$100+\frac{110}{1+r}=\frac{1}{1+r}+\frac{232}{(1+r)^2}$$

$$100(1+r)^2+110(1+r)-(1+r)-232=0$$

$$100+200r+100r^2+110+110r-1-r-232=0$$

$$100r^2+309r-23=0$$

Can you proceed using quadratic formula?

Answer 2

To simplify the problem, let $A = 1 + r.$ The equation becomes $$100 + 110A^{-1} = A^{-1} + 232A^{-2}.$$

Multiply both sides by $A^{2}$ to get $$100A^{2} + 110A = A + 232$$ $$100A^{2} + 109A - 232 = 0.$$

I leave you to solve the quadratic for $A$ and find the corresponding solutions for $r.$