Solve x(1+ay)^2=11025(1+a^2)(1)+11*25

$$x ( 1 + a y ) ^ { 2 } = 11025 ( 1 + a ^ { 2 } ) ( 1 ) + 11 \cdot 25$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{25\left(441a^{2}+452\right)}{\left(ay+1\right)^{2}}\text{, }&y=0\text{ or }a\neq -\frac{1}{y}\\x\in \mathrm{C}\text{, }&\left(y=\frac{21\sqrt{113}i}{226}\text{ and }a=\frac{2\sqrt{113}i}{21}\right)\text{ or }\left(y=-\frac{21\sqrt{113}i}{226}\text{ and }a=-\frac{2\sqrt{113}i}{21}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(1+ay\right)^{2}.

$$x\left(1+2ay+a^{2}y^{2}\right)=11025\left(1+a^{2}\right)\times 1+11\times 25$$

Use the distributive property to multiply x by 1+2ay+a^{2}y^{2}.

$$x+2xay+xa^{2}y^{2}=11025\left(1+a^{2}\right)\times 1+11\times 25$$

Multiply 11025 and 1 to get 11025.

$$x+2xay+xa^{2}y^{2}=11025\left(1+a^{2}\right)+11\times 25$$

Use the distributive property to multiply 11025 by 1+a^{2}.

$$x+2xay+xa^{2}y^{2}=11025+11025a^{2}+11\times 25$$

Multiply 11 and 25 to get 275.

$$x+2xay+xa^{2}y^{2}=11025+11025a^{2}+275$$

Add 11025 and 275 to get 11300.

$$x+2xay+xa^{2}y^{2}=11300+11025a^{2}$$

Combine all terms containing x.

$$\left(1+2ay+a^{2}y^{2}\right)x=11300+11025a^{2}$$

The equation is in standard form.

$$\left(a^{2}y^{2}+2ay+1\right)x=11025a^{2}+11300$$

Divide both sides by 1+2ay+a^{2}y^{2}.

$$\frac{\left(a^{2}y^{2}+2ay+1\right)x}{a^{2}y^{2}+2ay+1}=\frac{11025a^{2}+11300}{a^{2}y^{2}+2ay+1}$$

Dividing by 1+2ay+a^{2}y^{2} undoes the multiplication by 1+2ay+a^{2}y^{2}.

$$x=\frac{11025a^{2}+11300}{a^{2}y^{2}+2ay+1}$$

Divide 11300+11025a^{2} by 1+2ay+a^{2}y^{2}.

$$x=\frac{25\left(441a^{2}+452\right)}{\left(ay+1\right)^{2}}$$

Solve for x

$$x=\frac{25\left(441a^{2}+452\right)}{\left(ay+1\right)^{2}}$$

$$y=0\text{ or }a\neq -\frac{1}{y}$$

Steps for Solving Linear Equation

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(1+ay\right)^{2}.

$$x\left(1+2ay+a^{2}y^{2}\right)=11025\left(1+a^{2}\right)\times 1+11\times 25$$

Use the distributive property to multiply x by 1+2ay+a^{2}y^{2}.

$$x+2xay+xa^{2}y^{2}=11025\left(1+a^{2}\right)\times 1+11\times 25$$

Multiply 11025 and 1 to get 11025.

$$x+2xay+xa^{2}y^{2}=11025\left(1+a^{2}\right)+11\times 25$$

Use the distributive property to multiply 11025 by 1+a^{2}.

$$x+2xay+xa^{2}y^{2}=11025+11025a^{2}+11\times 25$$

Multiply 11 and 25 to get 275.

$$x+2xay+xa^{2}y^{2}=11025+11025a^{2}+275$$

Add 11025 and 275 to get 11300.

$$x+2xay+xa^{2}y^{2}=11300+11025a^{2}$$

Combine all terms containing x.

$$\left(1+2ay+a^{2}y^{2}\right)x=11300+11025a^{2}$$

The equation is in standard form.

$$\left(a^{2}y^{2}+2ay+1\right)x=11025a^{2}+11300$$

Divide both sides by 1+2ay+a^{2}y^{2}.

$$\frac{\left(a^{2}y^{2}+2ay+1\right)x}{a^{2}y^{2}+2ay+1}=\frac{11025a^{2}+11300}{a^{2}y^{2}+2ay+1}$$

Dividing by 1+2ay+a^{2}y^{2} undoes the multiplication by 1+2ay+a^{2}y^{2}.

$$x=\frac{11025a^{2}+11300}{a^{2}y^{2}+2ay+1}$$

Divide 11300+11025a^{2} by 1+2ay+a^{2}y^{2}.

$$x=\frac{25\left(441a^{2}+452\right)}{\left(ay+1\right)^{2}}$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{-xy+5\sqrt{452xy^{2}+441x-4983300}}{xy^{2}-11025}\text{; }a=-\frac{xy+5\sqrt{452xy^{2}+441x-4983300}}{xy^{2}-11025}\text{, }&y=0\text{ or }x\neq \frac{11025}{y^{2}}\\a=-\frac{x-11300}{2xy}\text{, }&x=\frac{11025}{y^{2}}\text{ and }y\neq 0\end{matrix}\right.$$

Solve for a

$$\left\{\begin{matrix}a=\frac{-xy+5\sqrt{452xy^{2}+441x-4983300}}{xy^{2}-11025}\text{; }a=-\frac{xy+5\sqrt{452xy^{2}+441x-4983300}}{xy^{2}-11025}\text{, }&\left(x\neq \frac{11025}{y^{2}}\text{ and }x\geq \frac{4983300}{452y^{2}+441}\right)\text{ or }\left(y=0\text{ and }x\geq 11300\right)\\a=-\frac{x-11300}{2xy}\text{, }&x=\frac{11025}{y^{2}}\text{ and }y\neq 0\end{matrix}\right.$$