Question

2. (2x-1)1/3 = (6x-5)1/3​

Answer 1

$\huge\sf\underline{Solution}$

$\dfrac{1(2x - 1)}{3} = \dfrac{1(6x - 5)}{3}$

$= > \dfrac{2x - 1}{3} = \dfrac{6x - 5}{3}$

Cross multiplying the terms:-

$= > 3(6x - 5) = 3(2x - 1)$

$= > 18x - 15 = 6x - 3$

$= > 18x - 6x = - 3 + 15$

$= > 12x = 12$

$= > x = \dfrac{12}{12}$

$= > x = 1$

Hence, the value of x is 1.

Verification:-

L.H.S

Putting the value of x

$= \dfrac{2x - 1}{3}$

$= \dfrac{ 2 \times 1 - 1}{3}$

$= \dfrac{2 - 1}{3}$

$= \dfrac{1}{3}$

Now, R.H.S

Putting the value of x

$= \dfrac{6x - 5}{3}$

$= \dfrac{6 \times 1 - 5}{3}$

$= \dfrac{6 - 5}{3}$

$= \dfrac{1}{3}$

Therefore,

L.H.S = R.H.S

Hence, proved.

Answer 2

Answer :-

x = 1

Solution :-

$\sf (2x - 1)\dfrac{1}{3} = (6x - 5) \dfrac{1}{3}$

$\sf (2x)\dfrac{1}{3} - 1( \dfrac{1}{3}) = (6x) \dfrac{1}{3} - 5(\dfrac{1}{3})$

$\sf \dfrac{2x}{3} - \dfrac{1}{3} = \dfrac{6x}{3} - \dfrac{5}{3}$

Transpose - 5/3 to LHS and 2x/3 to RHS

$\sf - \dfrac{1}{3} + \dfrac{5}{3} = \dfrac{6x}{3} - \dfrac{2x}{3}$

$\sf \dfrac{ - 1 + 5}{3} = \dfrac{6x - 2x}{3}$

$\sf \dfrac{4}{3} = \dfrac{4x}{3}$

Cancelling 3 on both sides

$\sf 4 = 4x$

$\sf x = \dfrac{4}{4}$

$\sf x = 1$

$\sf \therefore \: x = 1$

Verification :-

$\sf (2x - 1)\dfrac{1}{3} = (6x - 5) \dfrac{1}{3}$

$\sf (2(1) - 1)\dfrac{1}{3} = (6(1) - 5) \dfrac{1}{3}$

$\sf (2 - 1)\dfrac{1}{3} = (6 - 5) \dfrac{1}{3}$

$\sf (1)\dfrac{1}{3} = (1) \dfrac{1}{3}$

$\sf \dfrac{1}{3} = \dfrac{1}{3}$