# How do you solve # 2/ (b-2) = b/ (b^2 - 3b +2) + b/ (2b - 2 )#?

##### 2 Answers

I can not spot where I have gone wrong. I was not expecting an 'undefined' solution.

Is the question correct?

#### Explanation:

We are looking for common denominators so we need to make them all of the same format type.

Consider

Factors of 2 are

Note that

Rewrite the given equation as:

Note that

......................................................................................

Lets make them all have the denominator of:

Multiply by 1 and you do not change the value. However, 1 comes in many forms.

What follows may be folded if the equation is wider than the page.

Multiply all of both sides by

Now we solve as a standard quadratic

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Both LHS and RHS are undefined !!!

No solution.

#### Explanation:

First, factor the denominators:

Rewrite so that all fractions have a common denominator:

Multiply the entire equation by the

We're now left with

Check the solution by substituting it back into the original equation:

#b=2#

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

#2/(2-2) = 2/(2^2-3(2)+2) + 2/(2(2)-2)#

#2/0 = 2/0 + 1#

This is undefined, so there is no solution.