Solve 2+by^2+(a+b)z^2=1

$$2 + b y ^ { 2 } + ( a + b ) z ^ { 2 } = 1$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{by^{2}+bz^{2}+1}{z^{2}}\text{, }&z\neq 0\\a\in \mathrm{R}\text{, }&b=-\frac{1}{y^{2}}\text{ and }y\neq 0\text{ and }z=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a+b by z^{2}.

$$2+by^{2}+az^{2}+bz^{2}=1$$

Subtract 2 from both sides.

$$by^{2}+az^{2}+bz^{2}=1-2$$

Subtract 2 from 1 to get -1.

$$by^{2}+az^{2}+bz^{2}=-1$$

Subtract by^{2} from both sides.

$$az^{2}+bz^{2}=-1-by^{2}$$

Subtract bz^{2} from both sides.

$$az^{2}=-1-by^{2}-bz^{2}$$

Reorder the terms.

$$az^{2}=-by^{2}-bz^{2}-1$$

The equation is in standard form.

$$z^{2}a=-by^{2}-bz^{2}-1$$

Divide both sides by z^{2}.

$$\frac{z^{2}a}{z^{2}}=\frac{-by^{2}-bz^{2}-1}{z^{2}}$$

Dividing by z^{2} undoes the multiplication by z^{2}.

$$a=\frac{-by^{2}-bz^{2}-1}{z^{2}}$$

Divide -by^{2}-bz^{2}-1 by z^{2}.

$$a=-\frac{by^{2}+bz^{2}+1}{z^{2}}$$

Solve for b

$$b=-\frac{az^{2}+1}{y^{2}+z^{2}}$$

$$z\neq 0\text{ or }y\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply a+b by z^{2}.

$$2+by^{2}+az^{2}+bz^{2}=1$$

Subtract 2 from both sides.

$$by^{2}+az^{2}+bz^{2}=1-2$$

Subtract 2 from 1 to get -1.

$$by^{2}+az^{2}+bz^{2}=-1$$

Subtract az^{2} from both sides.

$$by^{2}+bz^{2}=-1-az^{2}$$

Reorder the terms.

$$by^{2}+bz^{2}=-az^{2}-1$$

Combine all terms containing b.

$$\left(y^{2}+z^{2}\right)b=-az^{2}-1$$

Divide both sides by y^{2}+z^{2}.

$$\frac{\left(y^{2}+z^{2}\right)b}{y^{2}+z^{2}}=\frac{-az^{2}-1}{y^{2}+z^{2}}$$

Dividing by y^{2}+z^{2} undoes the multiplication by y^{2}+z^{2}.

$$b=\frac{-az^{2}-1}{y^{2}+z^{2}}$$

Divide -az^{2}-1 by y^{2}+z^{2}.

$$b=-\frac{az^{2}+1}{y^{2}+z^{2}}$$