Solve 25/x^2+36+4/b^2=1

$$\frac { 25 } { x ^ { 2 } + 36 } + \frac { 4 } { b ^ { 2 } } = 1$$

Solve for b (complex solution)

$$b=-2i\left(-x^{2}-11\right)^{-\frac{1}{2}}\sqrt{x^{2}+36}$$

$$b=2i\left(-x^{2}-11\right)^{-\frac{1}{2}}\sqrt{x^{2}+36}\text{, }x\neq -6i\text{ and }x\neq 6i\text{ and }x\neq -\sqrt{11}i\text{ and }x\neq \sqrt{11}i$$

Solve for x (complex solution)

$$x=-i\left(4-b^{2}\right)^{-\frac{1}{2}}\sqrt{144-11b^{2}}$$

$$x=i\left(4-b^{2}\right)^{-\frac{1}{2}}\sqrt{144-11b^{2}}\text{, }b\neq 0\text{ and }b\neq 2\text{ and }b\neq -2$$

Solve for b

$$b=2\sqrt{\frac{x^{2}+36}{x^{2}+11}}$$

$$b=-2\sqrt{\frac{x^{2}+36}{x^{2}+11}}$$

Solve for x

$$x=\sqrt{-\frac{144-11b^{2}}{4-b^{2}}}$$

$$x=-\sqrt{-\frac{144-11b^{2}}{4-b^{2}}}\text{, }\left(b\geq -\frac{12\sqrt{11}}{11}\text{ and }b<-2\right)\text{ or }\left(b>2\text{ and }b\leq \frac{12\sqrt{11}}{11}\right)$$