# Solving this system

I have been using substitution as the method to solve this. It doesn't work and I also do some trial-and-error but it takes a lot of time before I found the solution. Now, Is there a better way of solving this system of equation? Thank You. $$\begin{cases} y =2000( 1 + 0.2x) \\ z= 2000(1.2)^x \\ z=2y\end{cases}$$

I am solving for $x$.

• Usually with exponential and polynomial equations, there are no general way of solving it. The best is to try small integer values, like $x=0$. Then, argue that only 1 solution exists, due to the shape of the graphs. – Calvin Lin Jun 27 '13 at 0:58

$$(1.2)^x=2+0.4x$$
From the following picture, we see that there are $2$ solutions, which are approximately $-3.73$ and $9.73$.
Substitution here is a great way to start: $$z = 2y \implies 2y = 2000(1.2)^x\implies y = 1000(1.2)^x$$ Now set this equal to the first equation, and you'll have an equation in one variable, $x$, for which you can then work to solve for $x$: $$y = 1000(1.2)^x = 2000(1 + 0.2 x)\iff (1.2)^x = 2 + 0.4 x$$