Solve 1/b_{2}-2/a+2=3/b+b^2

$$\frac { 1 } { b _ { 2 } } - \frac { 2 } { a + 2 } = \frac { 3 } { b + b ^ { 2 } }$$

Solve for a

$$a=-\frac{2\left(3b_{2}+b_{2}b^{2}-b+bb_{2}-b^{2}\right)}{3b_{2}-b-b^{2}}$$

$$b_{2}\neq \frac{b\left(b+1\right)}{3}\text{ and }b_{2}\neq 0\text{ and }b\neq -1\text{ and }b\neq 0$$

Steps for Solving Linear Equation

Variable a cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by bb_{2}\left(a+2\right)\left(b+1\right), the least common multiple of b_{2},a+2,b+b^{2}.

$$b\left(a+2\right)\left(b+1\right)-bb_{2}\left(b+1\right)\times 2=b_{2}\left(a+2\right)\times 3$$

Use the distributive property to multiply b by a+2.

$$\left(ba+2b\right)\left(b+1\right)-bb_{2}\left(b+1\right)\times 2=b_{2}\left(a+2\right)\times 3$$

Use the distributive property to multiply ba+2b by b+1.

$$ab^{2}+ba+2b^{2}+2b-bb_{2}\left(b+1\right)\times 2=b_{2}\left(a+2\right)\times 3$$

Use the distributive property to multiply bb_{2} by b+1.

$$ab^{2}+ba+2b^{2}+2b-\left(b_{2}b^{2}+bb_{2}\right)\times 2=b_{2}\left(a+2\right)\times 3$$

Use the distributive property to multiply b_{2}b^{2}+bb_{2} by 2.

$$ab^{2}+ba+2b^{2}+2b-\left(2b_{2}b^{2}+2bb_{2}\right)=b_{2}\left(a+2\right)\times 3$$

To find the opposite of 2b_{2}b^{2}+2bb_{2}, find the opposite of each term.

$$ab^{2}+ba+2b^{2}+2b-2b_{2}b^{2}-2bb_{2}=b_{2}\left(a+2\right)\times 3$$

Use the distributive property to multiply b_{2} by a+2.

$$ab^{2}+ba+2b^{2}+2b-2b_{2}b^{2}-2bb_{2}=\left(b_{2}a+2b_{2}\right)\times 3$$

Use the distributive property to multiply b_{2}a+2b_{2} by 3.

$$ab^{2}+ba+2b^{2}+2b-2b_{2}b^{2}-2bb_{2}=3b_{2}a+6b_{2}$$

Subtract 3b_{2}a from both sides.

$$ab^{2}+ba+2b^{2}+2b-2b_{2}b^{2}-2bb_{2}-3b_{2}a=6b_{2}$$

Subtract 2b^{2} from both sides.

$$ab^{2}+ba+2b-2b_{2}b^{2}-2bb_{2}-3b_{2}a=6b_{2}-2b^{2}$$

Subtract 2b from both sides.

$$ab^{2}+ba-2b_{2}b^{2}-2bb_{2}-3b_{2}a=6b_{2}-2b^{2}-2b$$

Add 2b_{2}b^{2} to both sides.

$$ab^{2}+ba-2bb_{2}-3b_{2}a=6b_{2}-2b^{2}-2b+2b_{2}b^{2}$$

Add 2bb_{2} to both sides.

$$ab^{2}+ba-3b_{2}a=6b_{2}-2b^{2}-2b+2b_{2}b^{2}+2bb_{2}$$

Combine all terms containing a.

$$\left(b^{2}+b-3b_{2}\right)a=6b_{2}-2b^{2}-2b+2b_{2}b^{2}+2bb_{2}$$

The equation is in standard form.

$$\left(b^{2}+b-3b_{2}\right)a=6b_{2}+2b_{2}b^{2}-2b+2bb_{2}-2b^{2}$$

Divide both sides by b+b^{2}-3b_{2}.

$$\frac{\left(b^{2}+b-3b_{2}\right)a}{b^{2}+b-3b_{2}}=\frac{6b_{2}+2b_{2}b^{2}-2b+2bb_{2}-2b^{2}}{b^{2}+b-3b_{2}}$$

Dividing by b+b^{2}-3b_{2} undoes the multiplication by b+b^{2}-3b_{2}.

$$a=\frac{6b_{2}+2b_{2}b^{2}-2b+2bb_{2}-2b^{2}}{b^{2}+b-3b_{2}}$$

Divide -2b^{2}+2bb_{2}-2b+2b_{2}b^{2}+6b_{2} by b+b^{2}-3b_{2}.

$$a=\frac{2\left(3b_{2}+b_{2}b^{2}-b+bb_{2}-b^{2}\right)}{b^{2}+b-3b_{2}}$$

Variable a cannot be equal to -2.

$$a=\frac{2\left(3b_{2}+b_{2}b^{2}-b+bb_{2}-b^{2}\right)}{b^{2}+b-3b_{2}}\text{, }a\neq -2$$

Solve for b (complex solution)

$$b=\frac{\sqrt{-\left(-a+2b_{2}-2\right)\left(12ab_{2}+a+22b_{2}+2\right)}+a-2b_{2}+2}{2\left(-a+2b_{2}-2\right)}$$

$$b=\frac{-\sqrt{-\left(-a+2b_{2}-2\right)\left(12ab_{2}+a+22b_{2}+2\right)}+a-2b_{2}+2}{2\left(-a+2b_{2}-2\right)}\text{, }a\neq -2\text{ and }b_{2}\neq 0\text{ and }b_{2}\neq \frac{a}{2}+1$$