Question 680334
{{{(2/3)n + (1/n) = 25/6}}}
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Multiply the entire equation by 6n to be rid of fractions
{{{6n(2/3)n + 6n(1/n) = 6n(25/6)}}}
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{{{4n^2 + 6 = 25n}}}
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{{{4n^2 - 25n + 6 = 0}}}
(4n - 1)(n - 6) = 0
n = 6
n = 1/4
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The number can be {{{highlight_green(6)}}} or {{{highlight_green((1/4))}}}
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Check
{{{(2/3)(6) + (1/6)}}} is {{{12/3 + 1/6 = 24/6 + 1/6 = 25/6}}} correct!
{{{(2/3)(1/4) + 4}}} is {{{2/12 + 4 = 2/12 + 48/12 = 50/12 = 25/6}}} correct!