Question

$\frac{3}{4} (7x - 1(2 x - \frac{1 - x}{2} )= x + \frac{3}{2}$
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Answer 1

Given Equation

$\rm\to \dfrac{3}{4}\bigg\{7x-1\bigg(2x-\dfrac{1-x}{2}\bigg)\bigg\}=x+\dfrac{3}{2}$

To Find the value of x

Now take

$\rm\to \dfrac{3}{4}\bigg\{7x-1\bigg(2x-\dfrac{1-x}{2}\bigg)\bigg\}=x+\dfrac{3}{2}$

Use BODMAS

$\rm\to \dfrac{3}{4}\bigg\{7x-1\bigg(\dfrac{2\times2x}{2}-\dfrac{1-x}{2}\bigg)\bigg\}=\dfrac{2x+3}{2}$

$\rm\to \dfrac{3}{4}\bigg\{7x-1\bigg(\dfrac{4x}{2}-\dfrac{1-x}{2}\bigg)\bigg\}=\dfrac{2x+3}{2}$

$\rm\to \dfrac{3}{4}\bigg\{7x-1\bigg(\dfrac{4x-(1-x)}{2}\bigg)\bigg\}=\dfrac{2x+3}{2}$

$\rm\to \dfrac{3}{4}\bigg\{7x-1\bigg(\dfrac{4x-1+x}{2}\bigg)\bigg\}=\dfrac{2x+3}{2}$

$\rm\to \dfrac{3}{4}\bigg\{7x-1\bigg(\dfrac{5x-1}{2}\bigg)\bigg\}=\dfrac{2x+3}{2}$

$\rm\to \dfrac{3}{4}\bigg\{7x-\dfrac{5x+1}{2}\bigg\}=\dfrac{2x+3}{2}$

$\rm\to \dfrac{3}{4}\bigg\{\dfrac{14x-5x+1}{2}\bigg\}=\dfrac{2x+3}{2}$

$\rm\to \dfrac{3}{4}\bigg\{\dfrac{9x+1}{2}\bigg\}=\dfrac{2x+3}{2}$

$\rm\to \dfrac{3}{2}\bigg\{\dfrac{9x+1}{2}\bigg\}={2x+3}$

$\rm\to \dfrac{27x+3}{4}={2x+3}$

$\rm\to27x+3=4(2x+3)$

$\rm\to 27x+3=8x+12$

$\rm\to 27x-8x=12-3$

$\rm\to 19x=9$

$\rm\to x=\dfrac{9}{19}$

Answer

$\rm\to x=\dfrac{9}{19}$