How to find the angle formed by two joined triangles when they are multiples? The problem is as follows:

The figure from below shows two triangles. Assume $PR=QS$. Using this
information find $x$.


The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&20^{\circ}\\
2.&18^{\circ}\\
3.&22^{\circ}\\
4.&24^{\circ}\\
4.&30^{\circ}\\
\end{array}$
What I attempted to do to solve this problem was to use the identity which relates the exterior sum of the two angles in a triangle. In other words:
$\angle PQS= 6x$
But that's it, I came stuck there. Since I found this problem in a section belonging to triangle congruency, I think it should be solved using such approach. Can someone help me here?. If so, what sort of the cases related with congruency of triangles is it?. Could it be that is side-angle-side?.
Please include a drawing in the solution as I dont know how to spot those in triangles.
 A: 
Construct point $A$ (as shown in the picture) such that $\overline{PS}=\overline{RA}$. Note that $$\overline{QS}=\overline{PR}=\overline{PS}+\overline{SR}=\overline{RA}+\overline{SR}=\overline{SA}$$ Therefore $$\angle SQA=\angle SAQ\implies 2x+\angle RQA=\angle RAQ\quad\quad(1)$$ Also note that $$\angle RQA+\angle RAQ=4x\quad\quad(2)$$
Solve system of equations and we have $$\angle RQA=x,\quad\angle RAQ=3x$$
Now since $\angle SPQ=\angle RAQ=3x$, we have $\overline{QP}=\overline{QA}$. Hence we can say that $$\triangle PQS\cong\triangle AQR\quad(S.A.S.)$$
so $$\overline{QS}=\overline{QR}\implies \angle QSR=\angle QRS=4x$$
Finally $$4x+4x+2x=180^{\circ}\implies \color{red}{x=18^{\circ}}$$
A: [![enter image description here][2]][2]
This figure satisfies the conditions. If $\angle QPR=\angle PQR=54^o$ then $\angle SQR=36^o$   then $ PR=QR=QS$. So correct option is $18^o$. Other options are wrong.
Moreover we can solve following equation:
$2x+4x+3x+\alpha=180$
Where $\alpha=\angle PQS$
Among numerous solutions only one set of solution; $(x, \alpha)=(18, 18)$ satisfies the condition
