Solve y+b=a^i+1/1-x^2

$$y + b = \frac { a ^ { i + 1 } } { 1 - x ^ { 2 } }$$

Solve for b

$$b=-\frac{-yx^{2}+y-a^{1+i}}{1-x^{2}}$$

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

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(x-1\right)\left(-x-1\right).

$$\left(x-1\right)\left(-x-1\right)y+\left(x-1\right)\left(-x-1\right)b=a^{i+1}$$

Use the distributive property to multiply x-1 by -x-1 and combine like terms.

$$\left(-x^{2}+1\right)y+\left(x-1\right)\left(-x-1\right)b=a^{i+1}$$

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

$$-x^{2}y+y+\left(x-1\right)\left(-x-1\right)b=a^{i+1}$$

Use the distributive property to multiply x-1 by -x-1 and combine like terms.

$$-x^{2}y+y+\left(-x^{2}+1\right)b=a^{i+1}$$

Use the distributive property to multiply -x^{2}+1 by b.

$$-x^{2}y+y-x^{2}b+b=a^{i+1}$$

Add x^{2}y to both sides.

$$y-x^{2}b+b=a^{i+1}+x^{2}y$$

Subtract y from both sides.

$$-x^{2}b+b=a^{i+1}+x^{2}y-y$$

Combine all terms containing b.

$$\left(-x^{2}+1\right)b=a^{i+1}+x^{2}y-y$$

The equation is in standard form.

$$\left(1-x^{2}\right)b=yx^{2}-y+a^{1+i}$$

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

$$\frac{\left(1-x^{2}\right)b}{1-x^{2}}=\frac{yx^{2}-y+a^{1+i}}{1-x^{2}}$$

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

$$b=\frac{yx^{2}-y+a^{1+i}}{1-x^{2}}$$

Solve for a

$$a=\left(|x-1||x+1||y+b|\right)^{\frac{1}{2}-\frac{1}{2}i}e^{\left(\frac{1}{2}+\frac{1}{2}i\right)arg(-\left(y+b\right)\left(x^{2}-1\right))+\pi n_{1}\left(-1-i\right)}$$

$$n_{1}\in \mathrm{Z}$$

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