Question

$\frac{2x - 1}{3} + 1 = \frac{x - 2}{3} + 2$
​

Answer 1

GIVEN :-

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

TO FIND :-

• Value of x.

SOLUTION :-

We have ,

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

Cross multiplying , we get..

$\\ \implies\sf \: \dfrac{2x - 1 + 1(3)}{3} = \dfrac{x - 2 + 2(3)}{3} \\ \\ \\ \implies\sf \: \dfrac{2x - 1 + 3}{3} = \dfrac{x - 2 + 6}{3} \\ \\ \\ \implies\sf \: \dfrac{2x + 2}{3} = \dfrac{x + 4}{3} \\ \\ \\ \implies\sf \: 3(2x + 2) = 3(x + 4)$

$\implies\sf \: 6x + 6 = 3x + 12 \\ \\ \\ \implies\sf \: 6x - 3x = 12 - 6 \\ \\ \\ \implies\sf \: 3x = 6 \\ \\ \\ \implies\sf \: x = \dfrac{6}{3} \\ \\ \\ \implies\underbrace{\boxed{ \mathfrak{x = 2}}}$

Hence , value of x is 2.

VERIFICATION :-

$\to\sf\dfrac{2x - 1}{3} + 1 = \dfrac{x - 2}{3} + 2 \\ \\ \\ \to \sf \: \dfrac{2(2) - 1}{3} + 1 = \dfrac{2 - 2}{3} + 2 \\ \\ \\ \to\sf \: \dfrac{4 - 1}{3} + 1 = \dfrac{0}{3} + 2 \\ \\ \\ \to \sf \: \cancel\dfrac{3}{3} + 1 = 0 + 2 \\ \\ \\ \to\sf \: 2 = 2 \: \: \: \: \: (verified)$

Answer 2

$\mathfrak{{\pmb{{\underline{To~solve}}:}}}$

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

$\mathfrak{{\pmb{{\underline{Solution}}:}}}$

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

$\sf\implies{{\dfrac{2x-1+3}{3}}={\dfrac{x-2+2(3)}{3}}~~~~~~~(LCM=3)}$

$\sf\implies{{\dfrac{2x+2}{\cancel{3}}}={\dfrac{x-2+6}{\cancel{3}}}}$

$\sf\implies{2x+2=x+4}$

$\sf\implies{2x-x=4-2}$

$\sf{~~~~~~~ {\blue{•~{\underline{\boxed{\sf{\pmb{x=2}}}}}}} }$

$\therefore\mathfrak{{\pmb{{\underline{Required~answer}}:}}}$

• x = 2

$\mathfrak{{\pmb{{\underline{Verification}}:}}}$

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

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

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

$\sf\implies{{\dfrac{3}{3}}+1={\dfrac{0}{3}}+2}$

$\sf\implies{1+1=0+2}$

$\sf\implies{{\blue{\underline{\boxed{\sf{\pmb{2=2}}}}}}}$

• Hence verified !