There are values A and B such that \(\[\frac{Bx-11}{x^2-7x+10}=\frac{A}{x-2}+\frac{3}{x-5}.\]\)Find A+B
+26367 There are values \(A\) and \(B\) such that \(\dfrac{Bx-11}{x^2-7x+10}=\dfrac{A}{x-2}+\dfrac{3}{x-5}\).
Find \(A+B\).
\(x^2-7x+10 = (x-2)(x-5)\)
\(\begin{array}{|lrcll|} \hline & \dfrac{Bx-11}{x^2-7x+10} &=& \dfrac{A}{x-2}+\dfrac{3}{x-5} \\\\ & \dfrac{Bx-11}{(x-2)(x-5)} &=& \dfrac{A}{x-2}+\dfrac{3}{x-5} \quad & | \quad \times (x-2)(x-5) \\\\ & Bx-11 &=& A(x-5) + 3(x-2) \\\\ \hline x=5: & 5B-11 &=& A(5-5) + 3(5-2) \\ & 5B-11 &=& 0 + 3*3 \\ & 5B-11 &=& 9 \\ & 5B &=& 9+11 \\ & 5B &=& 20 \\ & B &=& \dfrac{20}{5} \\ & \mathbf{ B} &=& \mathbf{4} \\ \hline x=2: & 2B-11 &=& A(2-5) + 3(2-2) \\ & 2B-11 &=& -3A + 0 \\ & 3A &=& 11-2B \quad &|\quad B=4 \\ & 3A &=& 11-2\cdot4 \\ & 3A &=& 11-8 \\ & 3A &=& 3 \quad &|\quad :3 \\ & \mathbf{A} &=& \mathbf{1} \\ \hline \end{array}\)
\(\mathbf{A+B} = 1+4 \mathbf{= 5}\)
