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Suppose a,b,c are positive reals such that

ab=2a + 2b,

ac= 3a + 3c,

bc = 4b + 4c

Find a+b+c

 

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 Sep 5, 2019

Best Answer 

 #1
avatar+25225 
+3

Suppose a,b,c are positive reals such that

ab=2a + 2b,

ac= 3a + 3c,

bc = 4b + 4c

Find a+b+c

 

\(\begin{array}{|lrcll|} \hline (1) & 2a + 2b &=& ab \\ & 2a-ab &=& -2b \\ & a(2-b) &=& -2b \\ & a &=& \dfrac{-2b}{2-b} \\ &\mathbf{ a } &=& \mathbf{ \dfrac{2b}{b-2} } \\ \hline (3) & 4b + 4c &=& bc \\ & 4c-bc &=& -4b \\ & c(4-b) &=& -4b \\ & c &=& \dfrac{-4b}{4-b} \\ &\mathbf{ c } &=& \mathbf{ \dfrac{4b}{b-4} } \\ \hline (2) & 3a + 3c &=& ac \\ & 3(a+c) &=& ac \\ & 3 \left(\dfrac{2b}{b-2}+\dfrac{4b}{b-4} \right) &=& \left(\dfrac{2b}{b-2}\right) \left(\dfrac{4b}{b-4}\right) \\ & 3 \left(\dfrac{2b(b-4)+4b(b-2)}{(b-2)(b-4)} \right) &=& \dfrac{8b^2}{(b-2)(b-4)} \\ & 3 \Big(2b(b-4)+4b(b-2)\Big) &=& 8b^2 \\ & 6b(b-4)+12b(b-2) &=& 8b^2 \quad | \quad : 2 \\ & 3b(b-4)+6b(b-2) &=& 4b^2 \\ & 3b^2-12b+6b^2-12b &=& 4b^2 \\ & 5b^2-24b &=& 0 \\ & b(5b-24) &=& 0 \\\\ & b &=& 0 \quad | \quad \text{no solution, while b is a positive real } \\\\ & 5b-24 &=& 0 \\ & \mathbf{b} &=& \mathbf{\dfrac{24}{5}} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{a} &=& \mathbf{ \dfrac{2b}{b-2} } \\\\ a &=& \dfrac{2\times\dfrac{24}{5}}{\dfrac{24}{5}-2} \\\\ a &=& \dfrac{48}{5\times\dfrac{(24-10)}{5}} \\\\ a &=& \dfrac{48}{14} \\\\ \mathbf{a} &=& \mathbf{\dfrac{24}{7}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{c} &=& \mathbf{ \dfrac{4b}{b-4} } \\\\ c &=& \dfrac{4\times\dfrac{24}{5}}{\dfrac{24}{5}-4} \\\\ c &=& \dfrac{96}{5\times\dfrac{(24-20)}{5}} \\\\ c &=& \dfrac{96}{4} \\\\ \mathbf{c} &=& \mathbf{24} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline a+b+c &=& \dfrac{24}{7} + \dfrac{24}{5} + 24 \\ &=& 24\times \left( \dfrac{1}{7} + \dfrac{1}{5} + 1 \right) \\ &=& 24\times \left( \dfrac{47}{35} \right) \\ \mathbf{a+b+c} &=& \mathbf{\dfrac{1128}{35}} \\ \hline \end{array} \)

 

laugh

 Sep 5, 2019
 #1
avatar+25225 
+3
Best Answer

Suppose a,b,c are positive reals such that

ab=2a + 2b,

ac= 3a + 3c,

bc = 4b + 4c

Find a+b+c

 

\(\begin{array}{|lrcll|} \hline (1) & 2a + 2b &=& ab \\ & 2a-ab &=& -2b \\ & a(2-b) &=& -2b \\ & a &=& \dfrac{-2b}{2-b} \\ &\mathbf{ a } &=& \mathbf{ \dfrac{2b}{b-2} } \\ \hline (3) & 4b + 4c &=& bc \\ & 4c-bc &=& -4b \\ & c(4-b) &=& -4b \\ & c &=& \dfrac{-4b}{4-b} \\ &\mathbf{ c } &=& \mathbf{ \dfrac{4b}{b-4} } \\ \hline (2) & 3a + 3c &=& ac \\ & 3(a+c) &=& ac \\ & 3 \left(\dfrac{2b}{b-2}+\dfrac{4b}{b-4} \right) &=& \left(\dfrac{2b}{b-2}\right) \left(\dfrac{4b}{b-4}\right) \\ & 3 \left(\dfrac{2b(b-4)+4b(b-2)}{(b-2)(b-4)} \right) &=& \dfrac{8b^2}{(b-2)(b-4)} \\ & 3 \Big(2b(b-4)+4b(b-2)\Big) &=& 8b^2 \\ & 6b(b-4)+12b(b-2) &=& 8b^2 \quad | \quad : 2 \\ & 3b(b-4)+6b(b-2) &=& 4b^2 \\ & 3b^2-12b+6b^2-12b &=& 4b^2 \\ & 5b^2-24b &=& 0 \\ & b(5b-24) &=& 0 \\\\ & b &=& 0 \quad | \quad \text{no solution, while b is a positive real } \\\\ & 5b-24 &=& 0 \\ & \mathbf{b} &=& \mathbf{\dfrac{24}{5}} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{a} &=& \mathbf{ \dfrac{2b}{b-2} } \\\\ a &=& \dfrac{2\times\dfrac{24}{5}}{\dfrac{24}{5}-2} \\\\ a &=& \dfrac{48}{5\times\dfrac{(24-10)}{5}} \\\\ a &=& \dfrac{48}{14} \\\\ \mathbf{a} &=& \mathbf{\dfrac{24}{7}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{c} &=& \mathbf{ \dfrac{4b}{b-4} } \\\\ c &=& \dfrac{4\times\dfrac{24}{5}}{\dfrac{24}{5}-4} \\\\ c &=& \dfrac{96}{5\times\dfrac{(24-20)}{5}} \\\\ c &=& \dfrac{96}{4} \\\\ \mathbf{c} &=& \mathbf{24} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline a+b+c &=& \dfrac{24}{7} + \dfrac{24}{5} + 24 \\ &=& 24\times \left( \dfrac{1}{7} + \dfrac{1}{5} + 1 \right) \\ &=& 24\times \left( \dfrac{47}{35} \right) \\ \mathbf{a+b+c} &=& \mathbf{\dfrac{1128}{35}} \\ \hline \end{array} \)

 

laugh

heureka Sep 5, 2019
 #2
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Awesome solution heureka!! 

Guest Sep 7, 2019

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