Question

solve the equation by transposition method and check the result​

Answer 1

Solve : -

$\bf \:  \dfrac{5}{4} (3x - 1) - 1(4x - \dfrac{2 - x}{3} ) = x + \dfrac{7}{3}$

$\sf \:  ⟼\dfrac{15x - 5}{4} \: - \: \dfrac{12x - (2 - x)}{3} = \dfrac{3x + 7}{3}$

$\sf \:  ⟼\dfrac{15x - 5}{4} - \dfrac{13x - 2}{3} = \dfrac{3x + 7}{3}$

$\sf \:  ⟼\dfrac{3(15x - 5) - \: 4(13x - 2)}{12} = \dfrac{3x + 7}{3}$

$\sf \:  ⟼\dfrac{45x - 15 - 52x + 8}{12} =\dfrac{3x + 7}{3}$

$\sf \:  ⟼\dfrac{ - 7x - 7}{12} = \dfrac{3x + 7}{3}$

☆ On cross multiplication, we get

$\sf \:  ⟼ \: - 21x - 21 = 36x + 84$

$\sf \:  ⟼ \: - 21x - 36x = 84 + 21$

$\sf \:  ⟼ - 57x = 105$

$\sf \:  ⟼x = - \dfrac{105}{57} = - \dfrac{35}{19}$

$\bf\implies \:x = - \dfrac{35}{19}$

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Verification:-

☆Consider LHS

$\bf \:  \dfrac{5}{4} (3x - 1) - 1(4x - \dfrac{2 - x}{3} )$

☆ On substituting the value of x, we get

$\sf \:  ⟼\dfrac{5}{4} \bigg(3 × \dfrac{ - 35}{19} - 1 \bigg) - \bigg(4 \times \dfrac{ - 35}{19} - \dfrac{2 - \dfrac{ - 35}{19} }{3} \bigg)$

$\sf \:  ⟼\dfrac{5}{4} \bigg(\dfrac{ - 105}{19} - 1 \bigg) - \bigg(\dfrac{ - 140}{19} - \dfrac{38 + 35}{19 \times 3} \bigg)$

$\sf \:  ⟼\dfrac{5}{4} \bigg(\dfrac{ - 105 - 19}{19} \bigg) - \bigg(\dfrac{ - 420 - 73}{57} \bigg)$

$\sf \:  ⟼\dfrac{5}{ \cancel{4}} \bigg(\dfrac{ \cancel{ - 124} \: \: \: ^ { - 31}}{19} \bigg) - \bigg(\dfrac{ - 493}{57} \bigg)$

$\sf \:  ⟼ - \dfrac{ 155}{19} + \dfrac{493}{57}$

$\sf \:  ⟼\dfrac{ - 465+ 493}{57}$

$\sf \:  ⟼\dfrac{28}{57}$

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☆ Consider RHS

$\sf \:  ⟼x + \dfrac{7}{3}$

On substituting the values of x, we get

$\sf \:  ⟼ - \dfrac{35}{19} + \dfrac{7}{3}$

$\sf \:  ⟼\dfrac{ - 105 + 133}{57} = \dfrac{28}{57}$

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$\bf\implies \:➦ LHS = RHS$

$\bf \:  \large\boxed{\bf\implies \:Hence,  \: Verified ✔}$

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Answer 2

Answer:

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