Question 454441
{{{(2a-1)/(2a-3) - (2a^2-5a+4)/(8a^2-22a+15) - (3a-2)/(4a-5)}}}
:
The middle denominator will factor to 
{{{(2a-1)/(2a-3) - (2a^2-5a+4)/((2a-3)(4a-5)) - (3a-2)/(4a-5)}}}
:
Obvious that the common denominator will be (2a-3)(4a-5), therefore we have
{{{((2a-1)(4a-5)-(2a^2-5a+4)-(3a-2)(2a-3))/((2a-4)(4a-5))}}}
:
FOIL
{{{((8a^2-10a-4a+5)-(2a^2-5a+4)-(6a^2-9a-4a+6))/((2a-4)(4a-5))}}}
:
{{{((8a^2-14a+5)-(2a^2-5a+4)-(6a^2-13a+6))/((2a-4)(4a-5))}}}
:
Remove brackets, a neg changes the signs
{{{(8a^2-14a+5-2a^2+5a-4-6a^2+13a-6)/((2a-4)(4a-5))}}}
:
Combine like terms, that gets rid of most of it, then (4a-5)'s cancel leaving:
{{{(8a^2-2a^2-6a^2-14a+5a+13a+5-4-6)/((2a-4)(4a-5))}}} = {{{(4a-5)/((2a-4)(4a-5))}}} = {{{1/(2a-4)}}}