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There exist real numbers A and B so that 1/(k(k+3)) = A/k + B/(k+3)
for all real numbers k other than 0 and -3. Find the ordered pair (A,B)

 Sep 18, 2019
 #1
avatar+109334 
+1

      1                       A                        B 

________  =       _____     +      _________

k ( k + 3)                 k                    k +  3

 

We can use  the method of partial fractions to solve this

Mutilply through by   k ( k +3)   and we have

 

1  =  A( k + 3)   + Bk        simplify

 

1 = (A + B)k  +  3A       equate terms   and we have this system

 

A + B    =   0 

3A  = 1    ⇒   A  = 1/3

 

Which means that B   =   -1/3  

 

So  (A, B)  =  (1/3,  -1/3 )

 

 

cool cool cool

 Sep 18, 2019
 #2
avatar+24378 
+2

There exist real numbers A and B so that \(\mathbf{ \dfrac{1}{k(k+3)} } = \mathbf{ \dfrac{A}{k} + \dfrac{B}{k+3} }\)
for all real numbers k other than 0 and -3. Find the ordered pair (A,B)

 

\(\begin{array}{|lrcll|} \hline &\mathbf{ \dfrac{1}{k(k+3)} } &=& \mathbf{ \dfrac{A}{k} + \dfrac{B}{k+3} } \\\\ &&& \boxed{k \neq 0 \text{ and } k \neq -3 } \\\\ &\dfrac{1}{k(k+3)} &=& \dfrac{A}{k} + \dfrac{B}{k+3} \quad | \quad \cdot k(k+3) \\\\ &\dfrac{k(k+3)}{k(k+3)} &=& \dfrac{Ak(k+3)}{k} + \dfrac{Bk(k+3)}{k+3} \\\\ &\mathbf{ 1 } &=& \mathbf{ A (k+3) + Bk } \\ \hline k=0: & 1 &=& A (0+3) + B\cdot 0 \\ & 1 &=& 3A \\ & \mathbf{A} &=& \mathbf{ \dfrac{1}{3} } \\ \hline k=-3: & 1 &=& A (-3+3) + B\cdot (-3) \\ & 1 &=& A \cdot 0 -3B \\ & 1 &=& -3B \\ & \mathbf{B} &=& \mathbf{ -\dfrac{1}{3} } \\ \hline \end{array} \)

 

laugh

 Sep 18, 2019

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