Solve for $x: \frac{( 3x+1)}{16}$​ + $\frac{( 2x−3)}{7}$​ = $\frac{( x+3)}{8}​ + \frac{( 3x−1)}{14}$.

Given: $\frac{( 3x+1)}{16}$​ + $\frac{( 2x−3)}{7}$​ = $\frac{( x+3)}{8}​ + \frac{( 3x−1)}{14}$.

To do: To solve the given expression for $x$

Solution:

$\frac{( 3 x + 1)}{16} + \frac{( 2 x - 3)}{7} = \frac{( x + 3)}{8} + \frac{( 3 x - 1)}{14}$

$\Rightarrow \frac{21 x + 7 + 32 x - 48}{112}  =  \frac{7 x + 21 + 12 x - 4}{56}$

$\Rightarrow \frac{53 x - 41}{112} = \frac{19 x + 17}{56}$     

$\Rightarrow 53 x - 41 = 38 x + 34$                           $[ (112 : 56)  = ( 2:1 )  ]$

$\Rightarrow 53 x-38 x = 34 + 41$

$\Rightarrow 15 x = 75$

$\Rightarrow x = \frac{75}{15}$

$\Rightarrow x = 5$

Updated on: 10-Oct-2022

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