$$\left\{\begin{array}{l} \ln(x+5)+y^2=10\\ x^2+y^2=9 \end{array}\right.$$
I have to solve this system of equations.
Let us assume:
$ x=ty$
The system becomes: $$\left\{\begin{array}{l} \ln(ty+5)+y^2=10\\ t^2y^2+y^2=9 \end{array}\right.$$
So:
$\dfrac{\ln(ty+5)}{t^2y^2} = \dfrac{10}{9}$
How can I go on?
We have $y=\pm \sqrt{9-x^2}$, so it is enough to find the solutions for $x$. Now $\log(x+5)=10-y^2=x^2+1$, so we have to solve the equation $$ \log(x+5)=x^2+1. $$ Here we can find help at MSE (e.g., a similar case is here).
As Dietrich Burde answered and Sujith Zis commented, eliminating $y^2$ from the second equation reduces to finding the zero's of $$f(x)=\log (x+5)-x^2-1$$ The derivative $$f'(x)=\frac{1}{x+5}-2 x$$ cancels for $$x_{\pm}=\frac{1}{2} \left(\pm 3 \sqrt{3}-5\right)$$ but $x_-$ has to be discarded since smaller than $-5$.
$$f(x_+)=-14+\frac{15 \sqrt{3}}{2}+\log \left(\frac{1}{2} \left(5+3 \sqrt{3}\right)\right)\approx 0.619244$$ and the second derivative test shows that this is a maximum.
Now, by inspection $$f(-1)=\log (4)-2\approx -0.613706$$ $$f(0)=\log (5)-1\approx 0.609438$$ $$f(1)=\log (6)-2\approx -0.208241$$ So, there are two roots : the first one between $-1$ and $0$, the second one between $0$ and $1$.
Since the second derivative $$f''(x)=-\frac{1}{(x+5)^2}-2$$ is always negative, thse are the only two roots for $x$.
At this point, you need a numerical method such as Newton to find their numerical values.
I am sure that you can take from here.
Edit
You could have a very good estimation of the roots if you expand the function as a Taylor series at $x=0$ (this is justified since we notice that the roots are significantly smaller than $5$); this will give $$f(x)=(\log (5)-1)+\frac{x}{5}-\frac{51 x^2}{50}+O\left(x^3\right)$$ Just by curiosity, plot the two functions on the same graph. From the solutions of the quadratic, I guess that one Newton iteration would be sufficient.
