Conditions for maximum power transfer in an AC circuit answer
Though your all-in-one approach is formally correct, as you already found out at your expenses, it practically takes you nowhere. You soon get lost into some hard boring math.

A much better way is to divide the problem in smaller easy to cope sub-sets.

But even before what you need is to get the right rig from your toolbox.  


  In this case  parallel-series resistance-reactance conversion is the the right way to do it.
  
  
  
  
  
  
  
  It can be easily proven that at one frequency, given \$Q=\frac{R_\text{p}}{X_\text{p}}=\frac{X_\text{s}}{R_\text{s}}\$ it holds true that
  $$R_\text{s}=R_\text{p}\,\frac{1}{Q^2+1}\quad\quad X_\text{s}=X_\text{p}\,\frac{Q^2}{Q^2+1}=Q\,R_\text{s}$$
  
  on a par with
  $$R_\text{p}=R_\text{s}(Q^2+1)\quad\quad
X_\text{p}=X_\text{s}\,\frac{Q^2+1}{Q^2}=\frac{R_\text{p}}{Q}$$


So, it's now time to try apply the above to your problem:


1) Transform the given load to parallel
$$Q=\frac{X_\text{s}}{R_\text{s}}=\frac{300\,\Omega}{200\,\Omega}=1.5$$
$$R_\text{p}=R_\text{s}\,(Q^2+1)=200\times(1.5^2+1)=650\,\Omega$$
$$X_\text{p}=\frac{R_\text{p}}{Q}=\frac{650\Omega}{1.5}=433.3\,\Omega$$


2) Now we shall find Q needed to match 650Ω resistance to 150Ω source once ZL and XC toghether will be reverted to series.
$$R_\text{s}=R_\text{p}\,\frac{1}{Q^2+1}\quad\Rightarrow\quad
Q=\sqrt{\frac{R_\text{p}}{R_\text{s}}-1}=\sqrt{\frac{650\,\Omega}{150\,\Omega}-1}=1.826$$

3) To achieve this we need a total parallel reactance as
$$Q=\frac{R_\text{p}}{X_\text{p}}\quad\Rightarrow\quad X_\text{p}=\frac{R_\text{p}}{Q}=\frac{650\,\Omega}{1.826}=356\,\Omega$$
which shall be achieved by parallelling unknown XC and -j433Ω, so
$$X_\text{c}=\frac{433.3\,\Omega\times 356\,\Omega}{433.3\,\Omega - 356\,\Omega}=1995\,\Omega$$



  in short \$X_\text{c}=1995\,\Omega\$


4) Back to series, everything needed is out there, we already know Q=1.826 and Rs=150Ω, so we just convert reactance
$$Q=\frac{X_\text{s}}{R_\text{s}}\quad\Rightarrow\quad X_\text{s}=Q\,R_\text{s}=1.826\times 150\,\Omega=274\,\Omega$$


5) Now it's really done, quite obviously 


  XL=274ohm 
  too, so to cancel series reactances.


6) Finally I'd rather go through a numerical check of our results



which, besides some rounding error, matches pretty well your specifications.

As you see, this dividi et impera approach turns a huge complex equation into just a few elementary operations.