Question

2t + 1/3 = t/3 + 1 please answer too fastttt​

Answer 1

Answer :-

$\sf t= \dfrac{2}{5}$

Solution :-

$\sf 2t + \dfrac{1}{3} = \dfrac{t}{3} + 1$

Transpose 1/3 to RHS [ + 1/3 becomes - 1/3]

$\sf 2t= \dfrac{t}{3} - \dfrac{1}{3} + 1$

Transpose t/3 to LHS [ + t/3 becomes - t/3]

$\sf 2t - \dfrac{t}{3} =- \dfrac{1}{3} + 1$

Taking LCM

$\sf \dfrac{2t(3)}{1(3)} - \dfrac{t}{3} =- \dfrac{1}{3} + \dfrac{1(3)}{1(3)}$

$\sf \dfrac{6t}{3} - \dfrac{t}{3} =- \dfrac{1}{3} + \dfrac{3}{3}$

$\sf \dfrac{6t - t}{3} = \dfrac{ - 1 + 3}{3}$

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

$\sf \dfrac{5t}{3} =\dfrac{2}{3}$

Transpose 3 to RHS [Division becomes Multiplication]

$\sf 5t= \dfrac{2}{3} \times 3$

$\sf 5t = \dfrac{6}{3}$

$\sf 5t=2$

Transpose 5 to RHS [Multiplication becomes division]

$\sf t= \dfrac{2}{5}$

$\sf \therefore t= \dfrac{2}{5}$

Verification :-

$\sf 2t + \dfrac{1}{3} = \dfrac{t}{3} + 1$

$\sf \implies 2( \dfrac{2}{5}) + \dfrac{1}{3} = \dfrac{\frac{2}{5} }{3} + 1$

$\sf \implies - \dfrac{4}{5} + \dfrac{1}{3} =\dfrac{2}{5} \div 3 + 1$

$\sf \implies \dfrac{4}{5} + \dfrac{1}{3} =\dfrac{2}{5} \times \dfrac{1}{3} + 1$

$\sf \implies \dfrac{4}{5} + \dfrac{1}{3} = \dfrac{2}{15} + 1$

$\sf \implies \dfrac{4(3)}{5(3)} + \dfrac{1(5)}{3(5)} = - \dfrac{1}{15} + \dfrac{1(15)}{1(15)}$

$\sf \implies \dfrac{12}{15} + \dfrac{5}{15} = \dfrac{2}{15} + \dfrac{15}{15}$

$\sf \implies \dfrac{12 + 5}{15} = \dfrac{2 + 15}{15}$

$\sf \implies \dfrac{17}{15} = \dfrac{17}{15}$

Answer 2

2t + 1/3 = t/3

2t - t/3 = - 1/3 + 1

Taking LCM

2t(3)/1(3) - t/3 = - 1/3 + 1

6t/3 - t/3 = - 1/3 + 1(3)/1(3)

(6t - t)/3 = - 1/3 + 3/3

5t/3 = ( - 1 + 3)/3

5t = - 2/3

5t = - 2/3 * 3

5t = -2

t = -2/5