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Solve for $x$:

$3-\frac{32}{20-\frac{16}{5-\frac{5}{x}} }=1$

When I see exercises like this one I panic because I can't figure out what to do with that variable x. What are the steps in solving an exercise like this one?

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It is quite difficult to read what you have posted.Do you know LaTex? :) – user31029 Jun 24 '12 at 12:43
My guess is that he means to solve $$3-\frac{32}{20-\frac{16}{5-\frac{5}{x}}}=1$$ – nullUser Jun 24 '12 at 13:00
@nullUser The answer should be 2. I don't know why Latex is needed in my question. It's just some calculations using PEMDAS. I have no problem with that but what confuses me is the variable x. – alistarmk Jun 24 '12 at 13:34
The solution is correct for the equation as parsed above. Is this the original form of the exercise in the book? Your notation is nonstandard and nullUser and I had to guess. – ncmathsadist Jun 24 '12 at 13:39
"The answer should be 2." Have you tried putting in 2 for $x$ to see whether it makes the left side equal the right side? If it doesn't, then 2 isn't the answer, whether it "should be" or not. – Gerry Myerson Jun 25 '12 at 6:32

Strip off the layers, one at a time, to bare the $x$. Begin with $$3 - {32\over{20 - {16\over 5 - 5/x}}}=1$$ Do a $3-$ on both sides to get $$ {32\over{20 - {16\over 5 - 5/x}}} = 2.$$ Multiply both sides by the denominator and you have $$40 - {32\over 5 - 5/x} = 32 $$ Now do a $40 -$ on both sides to obtain $${32\over 5 - 5/x} = 8.$$ Multiply both sides by the denominator and factor out the 5 to get $$32 = 8\cdot 5(1 - 1/x). $$ Divide by 40 and we get $${4\over 5} = 1 - 1/x.$$ Do a $1-$ on both sides to see that $1/x = 1/5$. We have $x = 5.$

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the answer should be 2. I looked at the answers section in this book. Here is the solution from the book: 32:[...]=3-1=2. I don't the solution but I hope it helps. – alistarmk Jun 24 '12 at 13:35
Well $x=2$ is certainly not the solution. Maybe the answers in the book are wrong... – fretty Jun 24 '12 at 14:55
@ncm, the last display isn't what you meant it to be. – Gerry Myerson Jun 25 '12 at 6:27
fixed. Thanks @GerryMyerson – ncmathsadist Jun 25 '12 at 13:37
$5$ is the solution to this problem. But maybe this transcription is not the problem actually in the book. Actually, all I see is Siminore's version, not the OP's "exactly as in the book" version. – GEdgar Jun 25 '12 at 14:04

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