Find the value of $n$ in $\dfrac{1}{3}(n+1)-1> \dfrac{1}{5}(n+4)$

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courageTyro

Asked: November 26, 20202020-11-26T20:51:31+00:00 2020-11-26T20:51:31+00:00In: Algebra

Find the value of $n$ in $\dfrac{1}{3}(n+1)-1> \dfrac{1}{5}(n+4)$

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Find the value of $y$ in $\dfrac{1}{3}(n+1)-1> \dfrac{1}{5}(n+4)$

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Sengemegz · Accepted Answer · 2020-12-02T11:41:34+00:00

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Sengemegz Moderator

2020-12-02T11:41:34+00:00Added an answer on December 2, 2020 at 11:41 am

$\dfrac{1}{3}(n+1)-1> \dfrac{1}{5}(n+4)$
multiply through by the LCM=(3\times 5=15)
$15\times \dfrac{1}{3}(n+1)-1> 15\times \dfrac{1}{5}(n+4)$
$5(n+1)> 3(n+4)$
$\Rightarrow 5n+5> 3n+12$
collect like terms
$5n-3n> 12-5$
$\Rightarrow 2n> 7$
divide both sides by 2
$\therefore n> \dfrac{7}{2}$

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