Question

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

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

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

Multiplying multiplying both sides by four the LCM of 4 and 2, we get

$3(7x - 1) - 8x + 4( \frac{1 - x}{2} ) = 4(x + \frac{3}{2}$

$⟹21x - 3 - 8x + 2 - 2x = 4x + 6$

$⟹21x - 8x - 2x - 3 + 2 = 4x + 6$

$⟹11x - 1 = 4x + 6$

$⟹11x - 4x = 6 + 1$

$⟹7x = 7$

$⟹\frac{7x}{7} = \frac{7}{7}$

$⟹x = 1$

$\sf{\underline{\underline{\pink{Thus,\:x\:=\:1\:is\:the\:solution\:of\:the\:given\: equation}}}}$

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${\huge{\underline{\small{\mathbb{\pink{</p><p>~AngelicCandy♡ :)}}}}}}$

Answer 2

━━━━━━━━━━━━━━━━━━━━━━━━━

$\huge\underline\bold\pink{AnSwEr}$

━━━━━━━━━━━━━━━━━━━━━━━━━

$\huge\underline\bold\orange{We\:Have}$

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

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

Multiplying multiplying both sides by four the LCM of 4 and 2, we get

$3(7x - 1) - 8x + 4( \frac{1 - x}{2} ) = 4(x + \frac{3}{2}$

$⟹21x - 3 - 8x + 2 - 2x = 4x + 6$

$⟹21x - 8x - 2x - 3 + 2 = 4x + 6$

$⟹11x - 1 = 4x + 6$

$⟹11x - 4x = 6 + 1$

$⟹7x = 7$

$⟹\frac{7x}{7} = \frac{7}{7}$

$⟹x = 1$

$\sf{\underline{\underline{\pink{Thus,\:x\:=\:1\:is\:the\:solution\:of\:the\:given\: equation}}}}$

━━━━━━━━━━━━━━━━━━━━━━━━━

${\huge{\underline{\small{\mathbb{\blue{HOPE\:HELP\:U\:BUDDY :)}}}}}}$

${\huge{\underline{\small{\mathbb{\pink{</p><p>~CandyFloss♡ :)}}}}}}$