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# Need help again D:

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There are values A and B such that (Bx-11)/(x^2-7x+10) = (A)/(x-2) + (3)/(x-5)
Find A+B.

May 28, 2019

#1
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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!

May 28, 2019

#1
0

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!

ElectricPavlov May 28, 2019
#2
+1

That's correct, thank you!

May 28, 2019