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

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 !!!

#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.