percentage problem for salary In $\;2002,\, 2003, \, \text{and}\; 2004\;$ the total income of Jerry  was $\$36,400$.   
His income increased by $15\%$ each year. What was his income in $2004$?
Any hints or solution will be welcome.
Thanks in advance.
 A: So let $x$ be his income in 2002. In 2003, his income was $x(1+0.15)$, in 2004, his income is $x(1+0.15)^2$. Therefore $x+x(1+0.15)+x(1+0.15)^2=36400$. Solve for x. Then the income in 2004 is just $x(1+0.15)^2$, where $x$ is what you just solved.
A: Let $I$ be the income earned in 2004 (what you want to know). Express everything in terms of this. e.g. Income in $2003$ is $\cfrac {I}{1.15}=\cfrac {20I}{23}$. That should reduce you to a simple equation for $I$.
A: $
\begin{align} I_0 & : \quad  \text{income earned in}\;2002.\tag{1} \\ \\
I_1 & =  1.15 I_0:\quad\text{income earned in}\; 2003.\tag{2} \\ \\
I_2 & = 1.15 I_1 = 1.15^2 I_0:\quad \text{income earned in} \; 2004.\tag{3}
\end{align}
$

$$\text{Income over $3$ years}:\;\;I_0 + I_1 + I_2 = \$36{,}400$$ $$ \iff I_0 + 1.15 I_0 + 1.15^2 = I_0\underbrace{(1 + 1.15 + 1.15^2)}_{\text{sum}\; =\; 3.4725}  = 36400 $$
Solve for $I_0$, the income earned in $2002$, and then compute $I_2$ (income earned in $2004),\,$ using your computed solution for $I_0$ and the relation given by $(3)$ above.
