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Answered by StormEyes
34

Solution!!

\sf \to \dfrac{2x}{3}-\dfrac{x-1}{6}+\dfrac{7x-1}{4}=2\dfrac{1}{6}

Convert into improper fraction.

\sf \to \dfrac{2x}{3}-\dfrac{x-1}{6}+\dfrac{7x-1}{4}=\dfrac{13}{6}

Taking the LCM.

\sf \to \dfrac{4(2x)-2(x-1)+3(7x-1)}{12}=\dfrac{13}{6}

Taking the denominator to the RHS.

\sf \to 4(2x)-2(x-1)+3(7x-1)=\dfrac{13}{6}\times 12

\sf \to 4(2x)-2(x-1)+3(7x-1)=13\times 2

\sf \to 4(2x)-2(x-1)+3(7x-1)=26

\sf \to 8x-2x+2+21x-3=26

\sf \to 27x-1=26

\sf \to 27x=26+1

\sf \to 27x=27

\sf \to x=\dfrac{\cancel{27}}{\cancel{27}}

\boxed{\sf x=1}

Verification.

Taking LHS

\sf \to \dfrac{2x}{3}-\dfrac{x-1}{6}+\dfrac{7x-1}{4}

\sf \to \dfrac{2x}{3}-\dfrac{x-1}{6}+\dfrac{7x-1}{4}

\sf \to \dfrac{2(1)}{3}-\dfrac{1-1}{6}+\dfrac{7(1)-1}{4}

\sf \to \dfrac{2}{3}-\dfrac{0}{6}+\dfrac{7-1}{4}

\sf \to \dfrac{2}{3}+\dfrac{6}{4}

\sf \to \dfrac{4(2)+3(6)}{12}

\sf \to \dfrac{8+18}{12}

\sf \to \dfrac{26}{12}

\sf \to \dfrac{13}{6}

\sf \to 2\dfrac{1}{6}

LHS = RHS

Hence, verified.

Answered by ItzDazzingBoy
0

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