Question

This question is from Linear Equation . Solve it .​

Answer 1

Given:

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

To find:-

• The value of x

Answer:-

Given that,

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

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

$\sf \longrightarrow\: \dfrac{21x - 3}{4} - \bigg \{ \dfrac{4x - (1 - x)}{2} \bigg \} = x + \dfrac{3}{2}$

$\sf \longrightarrow\: \dfrac{21x - 3}{4} - \bigg \{ \dfrac{4x - 1 + x}{2} \bigg \} = x + \dfrac{3}{2}$

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

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

$\sf \longrightarrow\: \dfrac{21x - 3 - 10x + 2}{4} = x + \dfrac{3}{2}$

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

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

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

$\sf \longrightarrow\: \dfrac{11x - 4x}{4} = \dfrac{2(3) + 1}{4}$

$\sf \longrightarrow\: \dfrac{7x}{4} = \dfrac{7}{4}$

$\sf \longrightarrow x = \dfrac{7}{4} \times \dfrac{4}{7}$

$\longrightarrow \underline{ \underline{ \sf { \green{x = 1}}}}$