Solve y=8/(frac{1{2}x+1)^2}*A

$$y = \frac { 8 } { ( \frac { 1 } { 2 } x + 1 ) ^ { 2 } } \cdot A$$

Solve for A

$$A=\frac{y\left(x+2\right)^{2}}{32}$$

$$x\neq -2$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{2}x+1\right)^{2}.

$$y=\frac{8}{\frac{1}{4}x^{2}+x+1}A$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{8}{\frac{1}{4}x^{2}+x+1}A=y$$

The equation is in standard form.

$$\frac{8}{\frac{x^{2}}{4}+x+1}A=y$$

Divide both sides by 8\left(\frac{1}{4}x^{2}+x+1\right)^{-1}.

$$\frac{\frac{8}{\frac{x^{2}}{4}+x+1}A\left(\frac{x^{2}}{4}+x+1\right)}{8}=\frac{y\left(\frac{x^{2}}{4}+x+1\right)}{8}$$

Dividing by 8\left(\frac{1}{4}x^{2}+x+1\right)^{-1} undoes the multiplication by 8\left(\frac{1}{4}x^{2}+x+1\right)^{-1}.

$$A=\frac{y\left(\frac{x^{2}}{4}+x+1\right)}{8}$$

Divide y by 8\left(\frac{1}{4}x^{2}+x+1\right)^{-1}.

$$A=\frac{y\left(x+2\right)^{2}}{32}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-4y^{-\frac{1}{2}}\sqrt{2A}-2\text{; }x=4\sqrt{2}y^{-\frac{1}{2}}\sqrt{A}-2\text{, }&A\neq 0\text{ and }y\neq 0\\x\neq -2\text{, }&y=0\text{ and }A=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=-4\sqrt{\frac{2A}{y}}-2\text{; }x=4\sqrt{\frac{2A}{y}}-2\text{, }&y>0\text{ and }A>0\\x=-4\sqrt{\frac{2A}{y}}-2\text{; }x=4\sqrt{\frac{2A}{y}}-2\text{, }&y<0\text{ and }A<0\\x\neq -2\text{, }&y=0\text{ and }A=0\end{matrix}\right.$$