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Find the value of x if  \(\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{x}}}}}=\frac{67}{96}\)

 

 

 

 

 

Need explanation, don't just plug this into the calculator...

 

Thanks to MaxWong for teaching me how to type in fractions using LaTeX

 Feb 8, 2019
edited by CalculatorUser  Feb 8, 2019
 #1
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+2

67 /96 =0.69791666666666666666666666666667  Take the reciprocal of this:

 

1.4328358208955223880597014925373    subtract 1 which is the first 1 in the denominator. Take the reciprocal of the fraction.

2.310344827586206896551724137931      subtract 2 which is the 2nd 2 in the denominator. Take the reciprocal of the fraction.

3.2222222222222222222222222222222    subtract 3 which is the 3rd 3 in the denominator. Take the reciprocal of the fraction.

4.5   subtract 4 which is the 4th digit in the denominator. Take the reciprocal of 0.5 which is = 2, which is the value of x in the denominator.

 Feb 8, 2019
 #2
avatar+101044 
+1

Thanks guest, I might not have thought of taking reciprocals without your help :)

 

\(\frac{67}{96}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{x}}}}}\\ \text{Take reciprocal of both sides}\\ \frac{96}{67}=\frac{1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{x}}}}}{1}\\~\\ \frac{96}{67}=1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{x}}}}\\~\\ \text{Subtract 1 both sides}\\ \frac{29}{67}=\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{x}}}}\\~\\ \text{Take reciprocal of both sides}\\ \frac{67}{29}=2+\frac{1}{3+\frac{1}{4+\frac{1}{x}}}\\~\\ \text{Subtract 2 both sides}\\ \frac{9}{29}=\frac{1}{3+\frac{1}{4+\frac{1}{x}}}\\~\\ \text{Take reciprocal of both sides}\\ \frac{29}{9}=3+\frac{1}{4+\frac{1}{x}}\\~\\ \text{Subtract 3 both sides}\\ \frac{2}{9}=\frac{1}{4+\frac{1}{x}}\\~\\ \text{Take reciprocal of both sides}\\ \frac{9}{2}=4+\frac{1}{x}\\~\\ \text{Subtract 4 both sides}\\ \frac{1}{2}=\frac{1}{x}\\~\\ \text{Take reciprocal of both sides}\\ x=\frac{1}{2}\)

 

That last line is incorrect.  since 1/x=1/2

x=2

.
 Feb 8, 2019
edited by Melody  Feb 8, 2019
 #3
avatar+22188 
+8

Find the value of \(\mathbf{ {\color{red}x}}\) if

 
\(\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}}}=\dfrac{67}{96}\)

 

\(\begin{array}{|rcll|} \hline \dfrac{67}{96} &=& \dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}}} \\ \left( 1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}} \right)67 &=& 96 \\\\ 67+\dfrac{67}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}} &=& 96 \\\\ \dfrac{67}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}} &=& 96-67 \\\\ \dfrac{67}{2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}}} &=& 29 \\\\ \left( 2+\dfrac{1}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}} \right)29 &=& 67 \\\\ 58+\dfrac{29}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}} &=& 67 \\\\ \dfrac{29}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}} &=& 67-58 \\\\ \dfrac{29}{3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}} &=& 9 \\\\ \left( 3+\dfrac{1}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}} \right)9&=& 29 \\\\ 27+\dfrac{9}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}&=& 29 \\\\ \dfrac{9}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}&=& 29-27 \\\\ \dfrac{9}{4+\dfrac{1}{\mathbf{ {\color{red}x}}}}&=& 2 \\\\ \left( 4+\dfrac{1}{\mathbf{ {\color{red}x}}} \right)2&=& 9 \\\\ 8+\dfrac{2}{\mathbf{ {\color{red}x}}}&=& 9 \\\\ \dfrac{2}{\mathbf{ {\color{red}x}}}&=& 9-8 \\\\ \dfrac{2}{\mathbf{ {\color{red}x}}}&=& 1 \\\\ \mathbf{2} & \mathbf{=} & \mathbf{ {\color{red}x} } \\ \hline \end{array}\)

 

laugh

 Feb 8, 2019
edited by heureka  Feb 8, 2019
 #4
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+2

Here's a neater method for getting to the result.

\(\begin{array}{c|ccccc} &1&2&3&4&x \\ \hline 0&1&2&7&30&\\ 1&1&3&10&43&&\\ \hline &1&2&3&4&x \end{array} \)

The 0 over 1 on the left reflects the fact that the fraction is less than 1, it would be 1 over 0 if the fraction were greater than 1.

The coefficients from the cf form the top and bottom rows, the two ones in the middle of the second column are just the first

coefficients. The remaining numbers in the middle (the 2 7 30 and 3 10 43 ) are calculated as " multiply the coefficient above or

below by the number below or above and to the left and then add the number further to the left of that ", so for example

 7 = 3*2 + 1, 30 = 4*7 + 2, 43 = 4*10 + 3.

Now ask what x must be in order to get the fraction 67/96.

If you take the numbers in the middle two rows and turn them into fractions, 1/1, 2/3, 7/10, 30/43, you find that these are the

simplified fractions coming from the cf truncated at its various points.

 

Tiggsy

 Feb 8, 2019
 #5
avatar+517 
+2

Thanks guys for the help!

 Feb 8, 2019

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