Question

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

Answer 1

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^2-3b+2`

Factors of 2 are `1 and 2`

Note that `(-1)xx(-2)=+2 and -1-2=-3` so we have:

`b^2-3b+2" "=" "(b-1)(b-2)`

Rewrite the given equation as:

`2/(b-2)=b/((b-1)(b-2))+b/(2b-2) " "larr checked`

Note that `2b-2" "=" "2(b-1)` so we now have:

`2/(b-2)=b/((b-1)(b-2))+b/(2(b-1))" "larr checked`
......................................................................................
Lets make them all have the denominator of: `2(b-1)(b-2)`

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

`color(green)([2/(b-2)color(red)(xx1)]=[b/((b-1)(b-2))color(red)(xx1)]+[b/(2(b-1))color(red)(xx1)] )`
`color(white)()`

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

`color(green)([2/(b-2)color(red)(xx(2(b-1))/(2(b-1)))]=[b/((b-1)(b-2))color(red)(xx2/2)])`
`color(green)(+[b/(2(b-1))color(red)(xx(b-2)/(b-2))] )" "larr checked`

`color(white)()`

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

Multiply all of both sides by `2(b-1)(b-2)` giving:

`4(b-1)=2b+b(b-2)`

`4b-4=cancel(2b)+b^2-cancel(2b)" " larr +2b-2b=0`

`b^2-4b+4=0`

Now we solve as a standard quadratic

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

`b=2+-sqrt(0)/2`

`b=2`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Check")`

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

`RHS->b/((b-1)(b-2))+b/(2b-2)`

`" "=2/((2-1)(2-2))+" "2/(2)`

`" "=" "color(red)(2/0)" "+" "1`
`" "color(red)(uarr)`
`color(red)(" Undefined")`

Both LHS and RHS are undefined !!!

Answer 2

No solution.

Explanation:

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

First, factor the denominators:

`2/(b-2) = b/((b-2)(b-1)) + b/(2(b-1))`

Rewrite so that all fractions have a common denominator:

`(color(red)2*2*color(blue)((b-1)))/(color(red)2(b-2)color(blue)((b-1))) = (color(red)2*b)/(color(red)2(b-2)(b-1)) + (bcolor(blue)((b-2)))/(2color(blue)((b-2))(b-1))`

Multiply the entire equation by the `2(b-2)(b-1)` to cancel the denominator:

`[(2*2*(b-1))/(2(b-2)(b-1)) = (2*b)/(2(b-2)(b-1)) + (b(b-2))/(2(b-2)(b-1))] * 2(b-2)(b-1)`

We're now left with

`2*2*(b-1) = (2*b) + b(b-2)`

`4(b-1)=2b+b(b-2) ->`simplify

`4b-4=2b+b^2-2b ->`expand

`0=b^2-4b+4 ->` rearrange so that all terms are on one side

`0=(b-2)^2 ->` factor

`0=b-2 ->` take the square root of both sides

`b=2 ->` add `2` to both sides

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.