Solve y+8=1/2(x+1)^2y(1)

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

Solve for y (complex solution)

$$y=-\frac{16}{1-2x-x^{2}}$$

$$x\neq \sqrt{2}-1\text{ and }x\neq -\left(\sqrt{2}+1\right)$$

Steps for Solving Linear Equation

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

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

Multiply \frac{1}{2} and 1 to get \frac{1}{2}.

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

Use the distributive property to multiply \frac{1}{2} by x^{2}+2x+1.

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

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

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

Subtract \frac{1}{2}x^{2}y from both sides.

$$y+8-\frac{1}{2}x^{2}y=xy+\frac{1}{2}y$$

Subtract xy from both sides.

$$y+8-\frac{1}{2}x^{2}y-xy=\frac{1}{2}y$$

Subtract \frac{1}{2}y from both sides.

$$y+8-\frac{1}{2}x^{2}y-xy-\frac{1}{2}y=0$$

Combine y and -\frac{1}{2}y to get \frac{1}{2}y.

$$\frac{1}{2}y+8-\frac{1}{2}x^{2}y-xy=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$\frac{1}{2}y-\frac{1}{2}x^{2}y-xy=-8$$

Combine all terms containing y.

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

The equation is in standard form.

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

Divide both sides by \frac{1}{2}-\frac{1}{2}x^{2}-x.

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

Dividing by \frac{1}{2}-\frac{1}{2}x^{2}-x undoes the multiplication by \frac{1}{2}-\frac{1}{2}x^{2}-x.

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

Divide -8 by \frac{1}{2}-\frac{1}{2}x^{2}-x.

$$y=-\frac{16}{1-2x-x^{2}}$$

Solve for y

$$y=-\frac{16}{1-2x-x^{2}}$$

$$x\neq \sqrt{2}-1\text{ and }x\neq -\sqrt{2}-1$$

Steps for Solving Linear Equation

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

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

Multiply \frac{1}{2} and 1 to get \frac{1}{2}.

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

Use the distributive property to multiply \frac{1}{2} by x^{2}+2x+1.

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

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

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

Subtract \frac{1}{2}x^{2}y from both sides.

$$y+8-\frac{1}{2}x^{2}y=xy+\frac{1}{2}y$$

Subtract xy from both sides.

$$y+8-\frac{1}{2}x^{2}y-xy=\frac{1}{2}y$$

Subtract \frac{1}{2}y from both sides.

$$y+8-\frac{1}{2}x^{2}y-xy-\frac{1}{2}y=0$$

Combine y and -\frac{1}{2}y to get \frac{1}{2}y.

$$\frac{1}{2}y+8-\frac{1}{2}x^{2}y-xy=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$\frac{1}{2}y-\frac{1}{2}x^{2}y-xy=-8$$

Combine all terms containing y.

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

The equation is in standard form.

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

Divide both sides by \frac{1}{2}-\frac{1}{2}x^{2}-x.

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

Dividing by \frac{1}{2}-\frac{1}{2}x^{2}-x undoes the multiplication by \frac{1}{2}-\frac{1}{2}x^{2}-x.

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

Divide -8 by \frac{1}{2}-\frac{1}{2}x^{2}-x.

$$y=-\frac{16}{1-2x-x^{2}}$$

Solve for x (complex solution)

$$x=\sqrt{2+\frac{16}{y}}-1$$

$$x=-\sqrt{2+\frac{16}{y}}-1\text{, }y\neq 0$$

Solve for x

$$\left\{\begin{matrix}x=-\sqrt{2+\frac{16}{y}}-1\text{; }x=\sqrt{2+\frac{16}{y}}-1\text{, }&y>0\\x=-\sqrt{2+\frac{16}{y}}-1\text{; }x=\sqrt{2+\frac{16}{y}}-1\text{, }&y\leq -8\end{matrix}\right.$$