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avatar+1438 

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

 

Find A+B

 

Thanks so much!

 Dec 19, 2017
 #1
avatar+101431 
+1

So we have that

 

[Bx  - 11 ] / [ x^2 -7x + 10]   =  A / [ x -2 ] + 3 / [x - 5]

 

Using partial fractions we can simplify this as

 

Bx  - 11   =  A[x - 5] + 3 [ x - 2]       simplify more

 

Bx  - 11  =  Ax - 5A  + 3x  - 6

 

Bx  - 11  = [ A + 3 ] x + [ -5A - 6 ]     equate coefficients

 

B  =  A +  3      

 

-11  = -5A - 6   ⇒  -5  =  -5A  ⇒    A  = 1

 

So....B  = 1 + 3   =  4

 

So....A + B  =   1  +  4  = 5

 

 

cool cool cool

 Dec 19, 2017
 #2
avatar+1438 
+2

Thanks!

AnonymousConfusedGuy  Dec 19, 2017

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