Question

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Answer 1

Question :

$\sf \dfrac{2x + 1}{10} - \dfrac{x-2}{6} = \dfrac{4-3x}{15}$

Answer :

Given :

• $\sf \dfrac{2x + 1}{10} - \dfrac{x-2}{6} = \dfrac{4-3x}{15}$

To Find :

• Value of x = ?

Solution :

$\sf : \implies \dfrac{2x + 1}{10} - \dfrac{x-2}{6} = \dfrac{4-3x}{15}$

$\sf : \implies \dfrac{(2x + 1) \times 6 }{10 \times 6} - \dfrac{(x-2)\times 10}{6\times 10} = \dfrac{4-3x}{15}$

$\sf : \implies \dfrac{12x + 6 }{60} - \dfrac{10x-20}{60} = \dfrac{4-3x}{15}$

$\sf : \implies \dfrac{(12x + 6) - (10x-20)}{60} = \dfrac{4-3x}{15}$

$\sf : \implies \dfrac{12x + 6 - 10x + 20}{60} = \dfrac{4-3x}{15}$

$\sf : \implies \dfrac{2x + 26}{60} = \dfrac{4-3x}{15}$

$\sf : \implies \dfrac{2(x + 13)}{60} = \dfrac{4-3x}{15}$

$\sf : \implies \dfrac{\cancel{2}(x + 13)}{\cancel{60}} = \dfrac{4-3x}{15}$

$\sf : \implies \dfrac{x + 13}{30} = \dfrac{4-3x}{15}$

By cross multiply :

$\sf : \implies (x + 13) \times 15 = (4-3x)\times 30$

$\sf : \implies 15x + 195 = 120 - 90x$

$\sf : \implies 15x + 195 - 120 + 90x = 0$

$\sf : \implies 105x + 75 = 0$

$\sf : \implies 105x = -75$

$\sf : \implies x = - \dfrac{75}{105}$

$\sf : \implies x = - \cancel{\dfrac{75}{105}}$

$\sf : \implies x = - \dfrac{5}{7}$

$\large \underline{\boxed{\bf{x = - \dfrac{5}{7}}}}$

$\large \bf Hence, \: value \: of \: x = - \dfrac{5}{7}$

Answer 2

Answer:

The value of x = -5/7

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