+9 9. THere are values A and B such that ((Bx-11)/(x^2-7x+10))=((A)/(x-2))+((3)/(x-5)). Find A + B.
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+1537 First, we try to "clear fractions" and see what we get:
x2−7x+10Bx−11=x−2A+x−53 Bx−11=A(x−5)+3(x−2) Expanding, we get: Bx−11=Ax−5A+3x−6 Bx−Ax=3x−5A−6 (B−A)x=−5A−6 If B−A=0, then we would need −5A−6=0, which means A=−56. But in this case, the equation (B−A)x=−5A−6 simplifies to 0=0, which means that the equation is true for any value of x. Since we know that A and B are unique values, we must have B−A=0. Therefore, we can divide both sides of the equation by B−A to get: x=B−A−5A−6 Setting x=2, we get: 2=B−A−5A−6 Solving for A, we get: A=52B−2 Substituting this value of A into the equation Bx−11=A(x−5)+3(x−2), we get: Bx−11=52B−2(x−5)+3(x−2) 5Bx−55=2Bx−10+3x−6 3Bx−55=3x−16 3Bx=3x−39 2Bx=−36 B=−x18 Since B is a unique value, we must have x=0. Therefore, B is a non-zero real number.
Setting x=5, we get: 5B−11=A(5−5)+3(5−2) 5B−11=3(5−2) 5B−11=9 5B=20 B=4 Substituting this value of B into the equation A=52B−2, we get: A=52(4)−2 A=56
Therefore, A+B=6/5+4=26/5.
