Answer
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Hint: In this question apply cross multiplication method to reach the solution of the question.
Given equation is
$\dfrac{{8x - 3}}{{3x}} = 2$
Now, 2 is also written as $\dfrac{2}{1}$
$ \Rightarrow \dfrac{{8x - 3}}{{3x}} = \dfrac{2}{1}$
Now apply cross multiplication we have
$ \Rightarrow \dfrac{{\left( {8x - 3} \right)}}{{\left( {3x} \right)}}$ = $\dfrac{2}{1}$
$
\left( {8x - 3} \right)1 = 3x\left( 2 \right) \\
\Rightarrow 8x - 3 = 6x \\
$
Now shifting the variables i.e. 6x to L.H.S and 3 to R.H.S
$
\Rightarrow 8x - 6x = 3 \\
\Rightarrow 2x = 3 \\
\Rightarrow x = \dfrac{3}{2} \\
$
So, this is the required solution of the given equation.
Note: In such types of questions the key concept we have to apply is cross multiplication and then shifting the variables to L.H.S and R.H.S respectively and after simplification, we will get the required answer.
Given equation is
$\dfrac{{8x - 3}}{{3x}} = 2$
Now, 2 is also written as $\dfrac{2}{1}$
$ \Rightarrow \dfrac{{8x - 3}}{{3x}} = \dfrac{2}{1}$
Now apply cross multiplication we have
$ \Rightarrow \dfrac{{\left( {8x - 3} \right)}}{{\left( {3x} \right)}}$ = $\dfrac{2}{1}$
$
\left( {8x - 3} \right)1 = 3x\left( 2 \right) \\
\Rightarrow 8x - 3 = 6x \\
$
Now shifting the variables i.e. 6x to L.H.S and 3 to R.H.S
$
\Rightarrow 8x - 6x = 3 \\
\Rightarrow 2x = 3 \\
\Rightarrow x = \dfrac{3}{2} \\
$
So, this is the required solution of the given equation.
Note: In such types of questions the key concept we have to apply is cross multiplication and then shifting the variables to L.H.S and R.H.S respectively and after simplification, we will get the required answer.
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