Solve rational equation

1. 2/b-2=b/b^2-3b+2 + b/2b-2

The way you typed it:

2/b-2=b/b^2-3b+2 + b/2b-2 reduces to
2/b-2=1/b-3b+2 + 1/b-2
multiply each term by b

2 - 2b = 1 - 3b^2 + 2b + 1 - 2b
3b^2 - 2b = 0
b(3b-2)
b = 0 or b = 2/3, but b ≠ 0
so b = 2/3

BUT

I have a strong feeling that you meant:

2/(b-2)=b/(b^2-3b+2) + b/(2b-2)
2/(b-2) = b/((b-1)(b-2)) + b/(2(b-1))
multiply each term by 2(b-1)(b-2)
4(b-1) = 2b + b(b-2)
4b - 4 = 2b + b^2 - 2b
b^2 -4b +4 = 0
(b-2)^2 = 0
b = 2,

but b ≠2 or else we are dividing by zero in the original question

so there is no solution.

To solve the rational equation 2/(b-2) = b/(b^2-3b+2) + b/(2b-2), we can follow these steps:

Step 1: Simplify the expressions on both sides of the equation by finding the common denominators.

The common denominator for the fractions on the right side of the equation is (b^2-3b+2)(2b-2). So, we need to rewrite the left side with the same denominator.

2/(b-2) = (2(b+2))/((b-2)(b+1))
= (2b+4)/(b^2-b-2)

Now, our equation becomes:

(2b+4)/(b^2-b-2) = b/(b^2-3b+2) + b/(2b-2)

Step 2: Combine the fractions on the right side of the equation using the same denominator.

(b/(b^2-3b+2) + b/(2b-2)) can be rewritten as:

(b(2b-2) + b(b^2-3b+2))/((b^2-3b+2)(2b-2))
= (2b^2 - 2b + b^3 - 3b^2 + 2b)/(b^3 - 3b^2 + 2b^2 - 6b + 4)
= (b^3 - b^2)/(b^3 - b^2 - 6b + 4)

So, our equation becomes:

(2b+4)/(b^2-b-2) = (b^3 - b^2)/(b^3 - b^2 - 6b + 4)

Step 3: Clear the fractions by multiplying both sides of the equation by the common denominator, which is (b^2-b-2)(b^3-b^2-6b+4).

(b^2-b-2)(b^3 - b^2 - 6b + 4)((2b+4)/(b^2-b-2)) = (b^2-b-2)(b^3 - b^2 - 6b + 4)((b^3 - b^2)/(b^3 - b^2 - 6b + 4))

Simplifying both sides by canceling out common factors, we get:

2b + 4 = b^3 - b^2

Step 4: Rearrange the equation to get all terms on one side and simplify if possible.

Rearranging, we have:

b^3 - b^2 - 2b - 4 = 0

Step 5: Solve the resulting polynomial equation.

Now, you can use various methods to solve this cubic equation, such as factoring, graphing, or using numerical methods. If factoring is not possible, one option is to use numerical methods like the Newton-Raphson method to approximate the solutions.