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# help

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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.

May 9, 2021

#1
-1

By partial fractions, A = 1 and B = 3.

May 9, 2021
#2
+32567
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As follows;

May 9, 2021
#3
+26213
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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.

$$\small{ \begin{array}{|lrcll|} \hline & \dfrac{Bx-11}{x^2-7x+10} &=&\dfrac{A}{x-2}+\dfrac{3}{x-5} \quad | \quad x^2-7x+10 =(x-2)(x-5) \\\\ & \dfrac{Bx-11}{(x-2)(x-5)} &=&\dfrac{A}{x-2}+\dfrac{3}{x-5} \quad | \quad * (x-2)(x-5)\\\\ & Bx-11 &=& A(x-5)+ 3(x-2) \\ & \mathbf{ Bx-A(x-5) } &=& \mathbf{ 3(x-2) + 11 } \\ \hline 1.~ x=0:& 0 +5A &=& -6+11 \\ & 5A &=& 5 \\ & \mathbf{A} &=& \mathbf{1} \\ \hline 2.~ x =5:& 5B - 0 &=& 9+11 \\ & 5B &=& 20 \\ & \mathbf{B} &=& \mathbf{4} \\ \hline \end{array} }$$

$$\mathbf{A+B = 5}$$

May 9, 2021
edited by heureka  May 9, 2021
edited by heureka  May 9, 2021