In the smaller triangle, let [tex]y[/tex] be the length of the smallest leg. Then the smallest leg in the larger triangle is [tex]14-y[/tex].
Within the scope of the smaller triangle, we have [tex]y[/tex] such that
[tex]x^2+y^2=13^2\implies y=\sqrt{13^2-x^2}[/tex]
Then within the larger the triangle, we would have
[tex]x^2+(14-y)^2=15^2\iff x^2+\left(14-\sqrt{13^2-x^2}\right)^2=15^2[/tex]
Now we can solve for [tex]x[/tex]:
[tex]x^2+14^2-28\sqrt{13^2-x^2}+(13^2-x^2)=15^2[/tex]
[tex]\implies x=12[/tex]
