Question

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Answer 1

Answer:

2x-1/3=x-2/3+1

2x-1/3=x-2+3/3

2x-1/3=x+1/3

by cross multiplication,

3(2x-1)=3(x+1)

cancelling 3 from both sides,

2x-1=x+1

x=2

Answer 2

Given :

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

To find : x

Solutions :

We are given linear equation in one variable.

Taking LCM on RHS, we get;

$\sf \implies \dfrac{2x - 1}{3} = \dfrac{ [ \: (x - 2) \times 1 \: ]+ (1 \times 3)}{3} \\ \\ \implies \: \sf \dfrac{2x - 1}{3} = \dfrac{ (\: x - 2 \: )+ (3)}{3} \\ \\ \sf \implies \dfrac{2x - 1}{3} = \dfrac{ \: x + 1}{3}$

(1/3) is on both side , that is LHS and RHS.

So they will cancel each other

$\sf \implies \: \dfrac{2x - 1}{ \not3} = \dfrac{ \: x + 1}{ \not3} \\ \\ \sf \implies \: 2x \: - \: \: 1 \: = \: x \: + \: 1 \\ \\ \sf \implies \: 2x \: - \: x \: = \: 1 \: + \: 1 \\ \\ \sf \implies \: \bold{x = 2}$

ANSWER :

x = 2