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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 $x$.

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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

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up vote 2 down vote accepted

Some simple substitutions lead to the equation:

$$(1.2)^x=2+0.4x$$

From the following picture, we see that there are $2$ solutions, which are approximately $-3.73$ and $9.73$.

enter image description here

There is no closed form for the solutions, but you can use Newton's method to achieve fairly accurate approximations.

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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 $x$: $$y = 1000(1.2)^x = 2000(1 + 0.2 x)\iff (1.2)^x = 2 + 0.4 x$$

With an exponential + polynomial equations, there is no tried-and-true way of solving the equation. However, you can try graphing each side of the final equation to see where the functions intersect.

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