Question

help again please maths.... linear equation​

Answer 1

Answer:

Taking LCM on both sides we get:

$\implies \dfrac{2n - n - 1}{2} = \dfrac{3 - n - 2}{3}\\\\\\\implies \dfrac{n-1}{2} = \dfrac{1-n}{3}\\\\\\\text{Cross Multiplying we get:}\\\\\\\implies 3(n-1) = 2(1-n)\\\\\implies 3n - 3 = 2 - 2n\\\\\implies 3n + 2n = 2 + 3 \\\\\implies 5n = 5\\\\\implies n = \dfrac{5}{5} = 1\\\\\implies \boxed{ \bf{n=1}}$

Hence the required value of 'n' is equal to 1.

Answer 2

Answer:-

$\\ \tt{:}\longrightarrow n-\dfrac{n-1}{2}=1-\dfrac{n-2}{3}$

$\\ \tt{:}\longrightarrow \dfrac{2n-(n-1)}{2}=\dfrac{3-(n-2)}{3}$

$\\ \tt{:}\longrightarrow \dfrac{2n-n+1}{2}=\dfrac{3-n+2}{3}$

$\\ \tt{:}\longrightarrow \dfrac{n+1}{2}=\dfrac{1-n}{3}$

• using Cross multiplication

$\\ \tt{:}\longrightarrow 3(n+1)=2(1-n)$

$\\ \tt{:}\longrightarrow 3n+3=2-2n$

$\\ \tt{:}\longrightarrow 3n+2n=2-3$

$\\ \tt{:}\longrightarrow 5n=-1$

$\\ \tt{:}\longrightarrow n=\dfrac{-1}{5}$