Question

$\frac{3t - 2}{4} - \frac{2t + 3}{3} = \frac{2}{3} - t$

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Answer 1

$\bf \underline{ \underline{\maltese\:Given} }$

$\sf \implies \dfrac{3t - 2}{4} - \dfrac{2t + 3}{3} = \dfrac{2}{3} - t$

$\bf \underline{ \underline{\maltese\:To \: find } }$

$\sf \implies Value \: of \: t = \: ?$

$\bf \underline{ \underline{\maltese\:Solution } }$

$\sf \implies Simplify \: \: \dfrac{3t - 2}{4} - \dfrac{2t + 3}{3}$

Find the common denominator.

$\sf \implies \dfrac{3t - 2}{4} \times \dfrac{3}{3} - \dfrac{2t + 3}{3} \times \dfrac{4}{4} = \dfrac{2}{3} - t$

$\sf \implies \dfrac{(3t - 2)3}{12} - \dfrac{(2t + 3)4}{12} = \dfrac{2}{3} - t$

Combine the numerators over common denominator.

$\sf \implies \dfrac{(3t - 2)3 - (2t + 3)4}{12} = \dfrac{2}{3} - t$

Simplify the numerator.

$\sf \implies \dfrac{9t - 6 - 8t + 12}{12} = \dfrac{2}{3} - t$

Subtract 8t from 9t.

$\sf \implies \dfrac{t - 6 - 12}{12} = \dfrac{2}{3} - t$

Subtract 12 from -6.

$\sf \implies \dfrac{t - 18}{12} = \dfrac{2}{3} - t$

Move all the terms containing t to the left side of the equation.

$\sf \implies \dfrac{t - 18}{12} + t= \dfrac{2}{3}$

Find the common denominator.

$\sf \implies \dfrac{t - 18}{12} + t \times \dfrac{12}{12} = \dfrac{2}{3}$

$\sf \implies \dfrac{t - 18}{12} + \dfrac{t \times 12}{12} = \dfrac{2}{3}$

Combine the numerators over common denominator.

$\sf \implies \dfrac{t - 18 + t \times 12}{12} = \dfrac{2}{3}$

$\sf \implies \dfrac{t - 18 +12 \times t}{12} = \dfrac{2}{3}$

Add t and 12t.

$\sf \implies \dfrac{13t - 18 }{12} = \dfrac{2}{3}$

Do cross multiplication,

$\sf \implies 3(13t - 18) = 2(12)$

$\sf \implies 39t - 54 = 24$

$\sf \implies 39t = 24 + 54$

$\sf \implies 39t = 78$

$\sf \implies t = \cancel\dfrac{78}{39}$

$\sf \implies t = 2$

Therefore,

$\sf \implies \underline{ \boxed { \sf Value \: of \: t = 2}}$

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$\bf \underline{ \underline{\maltese\:Verification } }$

To verify the answer just write 2 in place of t.

$\sf \implies \dfrac{3t - 2}{4} - \dfrac{2t + 3}{3} = \dfrac{2}{3} - t$

$\sf \implies \dfrac{3(2)- 2}{4} - \dfrac{2(2) + 3}{3} = \dfrac{2}{3} - 2$

$\sf \implies \dfrac{6- 2}{4} - \dfrac{4 + 3}{3} = \dfrac{2}{3} - \dfrac{2}{1}$

$\sf \implies \cancel\dfrac{4}{4} - \dfrac{7}{3} = \dfrac{2}{3} - \dfrac{2}{1}$

$\sf \implies 1 - \dfrac{7}{3} = \dfrac{2}{3} - \dfrac{2 \times 3 = 6}{1 \times 3 =3 }$

$\sf \implies \dfrac{3}{ 3} - \dfrac{7}{3} = \dfrac{2}{3} - \dfrac{6}{3}$

$\sf \implies \dfrac{ 3 - 7}{ 3}= \dfrac{2 - 6}{3}$

$\sf \implies \dfrac{ - 4}{ 3}= \dfrac{ - 4}{ 3}$

LHS and RHS are equal.

Hence verified !

Answer 2

Answer:

refer to the above attachment

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