jmc

Adding $\sqrt{25-x^2}$ to both sides gives $$\sqrt{49-x^2}=3+\sqrt{25-x^2}.$$Then, squaring both sides, we get $$49-x^2=9+6\sqrt{25-x^2}+(25-x^2),$$so $$15=6\sqrt{25-x^2}.$$Thus, $\sqrt{25-x^2}=\frac{15}{6}=\frac{5}{2}.$ Instead of solving for $x$ from here, we notice that $$\sqrt{49-x^2}=3+\sqrt{25-x^2}=3+\frac{5}{2}=\frac{11}{2}.$$Thus, $$\sqrt{49-x^2}+\sqrt{25-x^2}=\frac{11}{2}+\frac{5}{2}=\boxed{8}.$$