$x+y=5$
$2x+y-3z=12$
I know that in order to solve three unknowns three equations are needed, so I'm unsure if this can be solved or if different techniques (apart from the usual eliminations/substitutions) are used to solve problems like these.
[Math] How to solve a system of two equations in three unknowns
algebra-precalculussystems of equations
Related Solutions
If I properly understood the whole question, you have data points $(x_i,y_i)$ and you want to find the best fit for a model $$y=a x^b+c$$ which is intrinsically nonlinear (because of parameter $b$). The typical issue is to generate "reasonable" estimates of parameters $a,b,c$ for starting the nonlinear regression.
One thing you can notice is that the problem is very simple if $b$ is fixed. So, what you can do is to consider the function $SSQ(b)$ ($SSQ$ standing for the sum of squares of the residuals) and to plot it for various values of $b$; this just involves a linear regression. $SSQ(b)$ will go through a minimum and here you have your starting values.
For illustration purposes, I used the data given in the link provided by James Harrison. The minimum of $SS(b)$ is close to $b\approx 0.4$ and for this value $a\approx-11$, $c\approx 78$. This is more than sufficient to start the full nonlinear regression and get $$y= 83.5733 -14.5781 x^{0.35447}$$
By the way, working your three equations, suceesive eliminations of $c$ and $a$ let you with a single equation in $b$ $$F(b)=f^b (i-k)+h^b (k-g)+j^b (g-i)=0$$ to be looked graphically and may be solved by Newton method starting from a "reasonable" guess according to $$b_{n+1}=b_n-\frac{f^{b_n} (i-k)+h^{b_n} (k-g)+j^{b_n} (g-i)}{f^{b_n} \log (f) (i-k)+h^{b_n} (k-g) \log (h)+j^{b_n} (g-i) \log (j)}$$ which can be made much more compact defining a few intermediate constants.
You can also notice that writing $$F(b)=(i-k)e^{b\log(f)}+ (k-g)e^{b\log(h)}+ (g-i)e^{b\log(j)}=0$$ $$F(b)=(i-k)+ (k-g)e^{b\log(h/f)}+ (g-i)e^{b\log(j/f)}=0$$ reduces to a quadratic equation if the points, used for estimation of $a,b,c$ are selected in such a way that $\log(h/f)=2\log(j/f)$ that is to say if your three points are such that $h=j^2 f$.
If you derived the system of the equations correctly, you then get a normal 3x3 system equation with first-degree set of variables. This is a common linear algebra problem and can be solved in a ton of ways.
First of all, express $\frac{1}{x} = i , \frac{1}{y} = j , \frac{1}{z} = k$
Then you will have the following :
$ i + j + k = 1, 4i + 3j + 2k = 16, 2i - 2j - 3k = 5 $
In order to show that these can be solved or not, you can apply the Gauss Method on the matrix of their coefficients. Be careful after that, on probable solutions that may not apply for the starting fractions ($\frac{1}{x}$etc )
Best Answer
This system of equations can be solved, it just doesn't have a unique solution.
We can try to solve it as if we had just 2 variables, $x$ and $y$.
Subtract the first equation from the second to get $x = 3z + 7.$
Then since $y = 5 - x$ we get $y = -3z - 2.$
We've now solved the system, we just have one solution for each possible choice of $z$.