solve steady state of simple product reaction
There are $3$ equations for $3$ unknowns $[E],[S],[ES]$ given by

$$\left\{\begin{array}{l}
k_f[E][S]=k_r[ES],\\
[E]+[ES]=E_0,\\
[S]+[ES]=S_0.
\end{array}\right.$$

We can use the last two equations to cancel $[E]=E_0-[ES]$ and $[S]=S_0-[ES]$ and plug them into the first equation to get a quadratic equation for $[ES]$ given by

$$k_f(E_0-[ES])(S_0-[ES])=k_r[ES].$$

Then we plug in numbers $k_r/k_f=1$, $E_0=10$, $S_0=5$ to obtain $[ES]=8\pm\sqrt{14}$. Since we have $[ES]\leq\min\{E_0,S_0\}=5$, the solution is $[ES]=8-\sqrt{14}\approx 4.26$.