Math, asked by kajal8198, 1 year ago

2. (2x-1)1/3 = (6x-5)1/3​

Answers

Answered by Rose08
2

\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.

Answered by Anonymous
3

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}

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