SOLUTION:
Case: Simiar traiagles
Method:
It is good practice to separate the triangles into the two identifiable similar triangles
Using similar angle scale ratio:
[tex]\begin{gathered} \frac{13}{x+3}=\frac{13+x}{x+19} \\ 13(x+19)=(x+3)(13+x) \\ 13x+247=13x+x^2+39+3x \\ x^2+3x+13x-13x+39-_247=0 \\ x^2+3x-208=0 \\ Solving\text{ }the\text{ }quadratics \\ x=13\text{ }or\text{ }x=-16 \end{gathered}[/tex]
x cannot be -16 because no sides can have negative magnitudes
Hence, x can only be 13.
Final answer:
x= 13
