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Suppose that a, b, and c are real numbers for which \(\begin{align*}a^2b^2 &= 28, \\b^2c^2 &= 21, \text{ and}\\a^2c^2 &= 27,\end{align*}\)

and a

michaelcai  Oct 12, 2017
edited by michaelcai  Oct 12, 2017
edited by michaelcai  Oct 12, 2017
edited by michaelcai  Oct 12, 2017
 #1
avatar+86859 
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What's the question  ?

 

 

cool cool cool

CPhill  Oct 12, 2017
 #2
avatar+598 
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It doesnt show my edits for some reason. Here is the end of the question: and 

 

EDIT: UHOQIJKPWHOJKMI IT WONT LET ME USE THE LESS THAN BUTTON

 

 

here is the end: and \(a

 

Find 3b.

 

EDIT: I HAVE TO WORD IT NOW AHHHGHHHHH ok, so a less than b is less than c. Find 3b.

michaelcai  Oct 12, 2017
edited by michaelcai  Oct 12, 2017
edited by michaelcai  Oct 12, 2017
 #3
avatar+598 
0

Solved it: : Noticing that left-hand sides are symmetric in $a$, $b$, and $c$ (in that any relabelling of the variables in one of the left-hand sides gives one of the others), we multiply all three equations to get $(a^2b^2c^2)^2=(28)(27)(21)$, which implies $a^2b^2c^2=\sqrt{(28)(27)(21)}=\sqrt{(4\cdot 7)(9\cdot 3)(3\cdot 7.2\cdot 3\dot3\cdot 7=126.$ Dividing this equation by each of the three original equations gives $c^2 = 9/2$, $a^2 = 6$, and $b^2=14/3$, respectively. Since $b

michaelcai  Oct 12, 2017

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