Question

process + check please​

Answer 1

Soluction :-

$\sf \: we \: have$

$\sf \: = \frac{3}{4} (7x - 1) - \: \left ( \begin{array}{ccc} 2x - \frac{1 - x}{2} \end{array}\right ) = x + \frac{3}{2}$

$\sf \: \: = \left ( \begin{array}{ccc} \frac{21x}{4} - \frac{3}{4} \end{array}\right ) - \: \left ( \begin{array}{ccc} \frac{4x - 1 + x}{2} \end{array}\right ) = x + \frac{3}{2}$

$\sf \: \: = \left ( \begin{array}{ccc} \frac{21x}{4} - \frac{3}{4} \end{array}\right ) - \: \left ( \begin{array}{ccc} \frac{5x - 1}{2} \end{array}\right ) = x + \frac{3}{2}$

$\sf \: \: = \left ( \begin{array}{ccc} \frac{21x}{4} - \frac{3}{4} \end{array}\right ) - \: \left ( \begin{array}{ccc} \frac{5x}{2} - \frac{1}{2} \end{array}\right ) = x + \frac{3}{2}$

$\sf \: = \frac{21x}{4} - \frac{3}{4} - \frac{5x}{2} + \frac{1}{2} = x + \frac{3}{2}$

$\sf \: = \frac{21x - 10x}{4} - \frac{3}{4} + \frac{1}{2} = x + \frac{3}{4}$

$\sf \: = \frac{11x}{4} - x = \frac{3}{4} - \frac{1}{2} + \frac{3}{2}$

$\sf \: = \: \frac{11x - 4x}{4} = \frac{3 - 2 + 6}{4}$

$\sf \: = \frac{7x}{4} = \frac{7}{4}$

$\sf \: = 1$

$\sf \fbox \red{ans \: = \: x = 1}$

checking :-

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

$\sf \: = \frac{3}{4} \: (7 \times 1-1)-(2 \times\: 1 \: - \frac{1-1}{2})$

$\sf \: = \frac{9}{2} \:- 2\: = \: \frac{5}{2}$

$= \frac{5}{2} = \frac{5}{2}$

$\sf \: = 1$

$\sf \fbox \red{ans \: = \: x = 1}$

Observe that the left-hand side is equal to the right-hand side.

Hence, x = 1 is a solution to the given equation.

hence verified!!