There are values A and B such that (Bx-11)/(x^2-7x+10) = (A)/(x-2) + (3)/(x-5)
Find A+B.
Let's re-arrange the right side of the equation to a common denominator
a / (x-2) * (x-5)/(x-5) + 3/((x-5) * (x-2)/(x-2)
(ax-5a)/(x^2-7x+10) + (3x-6)/(x^2-7x+10) and combine the numerators
(ax+3x -5a-6)/(x^2-7x+10) now put the left side of the equation back in
(bx-11)/(x^2-7x+10) = (ax+3x-5a-6)/(x^2-7x+10) Note the L and R sides have the same denominator
so just equate the NUMERATORS
bx-11 = ax+3x-5a-6 now equate the components
bx = ax+3x and -11 = -5a-6
b = a+3 and -5=-5a so a =1
b=1+3 = 4
a+b = 5 That is the methodolology......you might want to check my work!
Let's re-arrange the right side of the equation to a common denominator
a / (x-2) * (x-5)/(x-5) + 3/((x-5) * (x-2)/(x-2)
(ax-5a)/(x^2-7x+10) + (3x-6)/(x^2-7x+10) and combine the numerators
(ax+3x -5a-6)/(x^2-7x+10) now put the left side of the equation back in
(bx-11)/(x^2-7x+10) = (ax+3x-5a-6)/(x^2-7x+10) Note the L and R sides have the same denominator
so just equate the NUMERATORS
bx-11 = ax+3x-5a-6 now equate the components
bx = ax+3x and -11 = -5a-6
b = a+3 and -5=-5a so a =1
b=1+3 = 4
a+b = 5 That is the methodolology......you might want to check my work!