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

Jul 13, 2017

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 ${b}^{2} - 3 b + 2$

Factors of 2 are $1 \mathmr{and} 2$

Note that $\left(- 1\right) \times \left(- 2\right) = + 2 \mathmr{and} - 1 - 2 = - 3$ so we have:

${b}^{2} - 3 b + 2 \text{ "=" } \left(b - 1\right) \left(b - 2\right)$

Rewrite the given equation as:

$\frac{2}{b - 2} = \frac{b}{\left(b - 1\right) \left(b - 2\right)} + \frac{b}{2 b - 2} \text{ } \leftarrow c h e c k e d$

Note that $2 b - 2 \text{ "=" } 2 \left(b - 1\right)$ so we now have:

$\frac{2}{b - 2} = \frac{b}{\left(b - 1\right) \left(b - 2\right)} + \frac{b}{2 \left(b - 1\right)} \text{ } \leftarrow c h e c k e d$
......................................................................................
Lets make them all have the denominator of: $2 \left(b - 1\right) \left(b - 2\right)$

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

$\textcolor{g r e e n}{\left[\frac{2}{b - 2} \textcolor{red}{\times 1}\right] = \left[\frac{b}{\left(b - 1\right) \left(b - 2\right)} \textcolor{red}{\times 1}\right] + \left[\frac{b}{2 \left(b - 1\right)} \textcolor{red}{\times 1}\right]}$
$\textcolor{w h i t e}{}$

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

$\textcolor{g r e e n}{\left[\frac{2}{b - 2} \textcolor{red}{\times \frac{2 \left(b - 1\right)}{2 \left(b - 1\right)}}\right] = \left[\frac{b}{\left(b - 1\right) \left(b - 2\right)} \textcolor{red}{\times \frac{2}{2}}\right]}$
$\textcolor{g r e e n}{+ \left[\frac{b}{2 \left(b - 1\right)} \textcolor{red}{\times \frac{b - 2}{b - 2}}\right]} \text{ } \leftarrow c h e c k e d$

$\textcolor{w h i t e}{}$

$\frac{4 \left(b - 1\right)}{2 \left(b - 1\right) \left(b - 2\right)} = \frac{2 b}{2 \left(b - 1\right) \left(b - 2\right)} + \frac{b \left(b - 2\right)}{2 \left(b - 1\right) \left(b - 2\right)}$

Multiply all of both sides by $2 \left(b - 1\right) \left(b - 2\right)$ giving:

$4 \left(b - 1\right) = 2 b + b \left(b - 2\right)$

$4 b - 4 = \cancel{2 b} + {b}^{2} - \cancel{2 b} \text{ } \leftarrow + 2 b - 2 b = 0$

${b}^{2} - 4 b + 4 = 0$

Now we solve as a standard quadratic

$b = \frac{+ 4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(1\right) \left(+ 4\right)}}{2 \left(1\right)}$

$b = 2 \pm \frac{\sqrt{0}}{2}$

$b = 2$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Check}}$

LHS->color(red)(2/(b-2))" "=" "color(red)(2/0 larr" Undefined")

$R H S \to \frac{b}{\left(b - 1\right) \left(b - 2\right)} + \frac{b}{2 b - 2}$

$\text{ "=2/((2-1)(2-2))+" } \frac{2}{2}$

$\text{ "=" "color(red)(2/0)" "+" } 1$
$\text{ } \textcolor{red}{\uparrow}$
$\textcolor{red}{\text{ Undefined}}$

Both LHS and RHS are undefined !!!

Jul 13, 2017

No solution.

#### Explanation:

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

First, factor the denominators:

$\frac{2}{b - 2} = \frac{b}{\left(b - 2\right) \left(b - 1\right)} + \frac{b}{2 \left(b - 1\right)}$

Rewrite so that all fractions have a common denominator:

$\frac{\textcolor{red}{2} \cdot 2 \cdot \textcolor{b l u e}{\left(b - 1\right)}}{\textcolor{red}{2} \left(b - 2\right) \textcolor{b l u e}{\left(b - 1\right)}} = \frac{\textcolor{red}{2} \cdot b}{\textcolor{red}{2} \left(b - 2\right) \left(b - 1\right)} + \frac{b \textcolor{b l u e}{\left(b - 2\right)}}{2 \textcolor{b l u e}{\left(b - 2\right)} \left(b - 1\right)}$

Multiply the entire equation by the $2 \left(b - 2\right) \left(b - 1\right)$ to cancel the denominator:

$\left[\frac{2 \cdot 2 \cdot \left(b - 1\right)}{2 \left(b - 2\right) \left(b - 1\right)} = \frac{2 \cdot b}{2 \left(b - 2\right) \left(b - 1\right)} + \frac{b \left(b - 2\right)}{2 \left(b - 2\right) \left(b - 1\right)}\right] \cdot 2 \left(b - 2\right) \left(b - 1\right)$

We're now left with

$2 \cdot 2 \cdot \left(b - 1\right) = \left(2 \cdot b\right) + b \left(b - 2\right)$

$4 \left(b - 1\right) = 2 b + b \left(b - 2\right) \to$simplify

$4 b - 4 = 2 b + {b}^{2} - 2 b \to$expand

$0 = {b}^{2} - 4 b + 4 \to$ rearrange so that all terms are on one side

$0 = {\left(b - 2\right)}^{2} \to$ factor

$0 = b - 2 \to$ take the square root of both sides

$b = 2 \to$ add $2$ to both sides

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

$b = 2$

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

$\frac{2}{2 - 2} = \frac{2}{{2}^{2} - 3 \left(2\right) + 2} + \frac{2}{2 \left(2\right) - 2}$

$\frac{2}{0} = \frac{2}{0} + 1$

This is undefined, so there is no solution.