How do you solve $\frac{1}{5 - \frac{1}{1 - \frac{1}{x}}} = \frac{2}{7}$ $\frac { 1} { 5- \frac { 1} { 1- \frac { 1} { x } } } = \frac { 2} { 7}$ ?

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Answer 1 · 2017-12-07T22:18:06

$x = 3$ $x=3$

This is kind of a mess, but as long as we take it step by step, we'll be fine

$\frac{1}{5 - \frac{1}{1 - \frac{1}{x}}} = \frac{2}{7}$ $1/(5-1/(1-1/(x)))=2/7$

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Let's make $1 - \frac{1}{x}$ $1-1/x$ one fraction with a common denominator

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$\frac{1}{5 - \frac{1}{\frac{x - 1}{x}}} = \frac{2}{7}$ $1/(5-1/((x-1)/x))=2/7$

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Now, let's change this mess of $\frac{1}{\frac{x - 1}{x}}$ $1/((x-1)/x)$ to $\frac{x}{x - 1}$ $x/(x-1)$

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$\frac{1}{5 - \frac{x}{x - 1}} = \frac{2}{7}$ $1/(5-x/(x-1))=2/7$

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Now, if we change $5$ $5$ to $\frac{5}{1}$ $5/1$ , we can give it the same denominator: $\frac{5 x - 5}{x - 1}$ $(5x-5)/(x-1)$

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That gives us

$\frac{1}{\frac{5 x - 5 - x}{x - 1}} = \frac{2}{7}$ $1/((5x-5-x)/(x-1))=2/7$

or

$\frac{x - 1}{4 x - 5} = \frac{2}{7}$ $(x-1)/(4x-5)=2/7$

See how much better this looks!! We are almost there

Let's clear the denominator

$7 x - 7 = 8 x - 10$ $7x-7=8x-10$

$x = 3$ $x=3$

There! We are done, good work

How do you solve 15−11−1x=27\frac { 1} { 5- \frac { 1} { 1- \frac { 1} { x } } } = \frac { 2} { 7}?