Solving the system $3.45e-x+4.65y=13$, $x^{e}+1.65y=4$

Question

I have this system \begin{align} 3.45e-x+4.65y&=13 \\ x^{e}+1.65y&=4 \end{align} But i can't determine what kind of problem this is to resolve it from mathway.

Answer 1

As you can guess, this is a non-linear system of equations. It is here trivial to eliminate $y$ so that you are left with a scalar non-linear equation in $x$. Here and in general there is no symbolic solution, the best you can do is to identify intervals with a sign change and use a bracketing numerical method to find root approximations.

Answer 2

Solving for $y$ and eliminating it from the system of equations, one is left with

$$\frac1{1.65}x^e+\frac1{4.65}x+C=0$$

where $C=\frac{13-3.45e}{4.65}-\frac4{1.65}$ is constant. Since the RHS is increasing in $x$ over $[0,\infty)$, there is a unique solution and it may be easily estimated as $x\approx1.34486025094$ using methods such as Newton's method.

Substituting this back in for $y$, one finds $y\approx1.0681264393$.