Question

I need a step by step explanation
please​

Answer 1

Answer:

2.5

Step-by-step explanation:

This is the answer.

Hope it is helpful

Answer 2

Given :

${\sf \dfrac{2}{3}(4x-1)-\bigg[2x-\dfrac{1+x}{3}\bigg]=\dfrac{1}{3}x+\dfrac{4}{3}}$

Solution :

${\Rightarrow \sf \dfrac{2}{3}(4x-1)-\bigg[2x-\dfrac{1+x}{3}\bigg]=\dfrac{1}{3}x+\dfrac{4}{3}}$

${\Rightarrow \sf \dfrac{2(4x-1)}{3}-\bigg[\dfrac{6x-1-x}{3}\bigg]=\dfrac{x}{3}+\dfrac{4}{3}}$

${\Rightarrow \sf \dfrac{8x-2}{3}-\bigg[\dfrac{5x-1}{3}\bigg]=\dfrac{x+4}{3}}$

${\Rightarrow \sf \bigg[\dfrac{8x-2-5x+1}{3}\bigg] =\dfrac{x+4}{3}}$

${\Rightarrow \sf \dfrac{3x-1}{3}=\dfrac{x+4}{3}}$

${\Rightarrow \sf \dfrac{3x-1}{3}-\bigg(\dfrac{x+4}{3}\bigg)=0}$

${\Rightarrow \sf \dfrac{3x-1-x-4}{3}=0}$

${\Rightarrow \sf \dfrac{2x-5}{3}=0}$

${\Rightarrow \sf 2x-5=0}\\\\{\Rightarrow 2x=5}\\\\{\Rightarrow \boxed{\sf x=\dfrac{5}{2}}}$

Verification :

${\Rightarrow \sf \dfrac{2}{3}(4x-1)-\bigg[2x-\dfrac{1+x}{3}\bigg]=\dfrac{1}{3}x+\dfrac{4}{3}}$

Substitute the value of x in place of x,

${\Rightarrow \sf \dfrac{2}{3}\Bigg(4\bigg(\dfrac{5}{2}\bigg)-1\Bigg)-\bigg[2\bigg(\dfrac{5}{2}\bigg)-\dfrac{1+\bigg(\dfrac{5}{2}\bigg)}{3}\bigg]=\dfrac{1}{3}\bigg(\dfrac{5}{2}\bigg)+\dfrac{4}{3}}$

${\Rightarrow\sf \dfrac{2}{3}\bigg(\dfrac{20}{2}-1\bigg)-\Bigg[\dfrac{10}{2}-\dfrac{\dfrac{2+5}{2}}{3}\Bigg]=\dfrac{5}{6}+\dfrac{4}{3}}$

${\Rightarrow\sf \dfrac{2}{3}(9)-\Bigg[5-\bigg(\dfrac{2+5}{2}\bigg)\bigg(\dfrac{1}{3}\bigg)\Bigg]=\dfrac{5}{6}+\dfrac{8}{6}}$

${\Rightarrow \sf \dfrac{18}{3}-\Bigg[5-\bigg(\dfrac{7}{2}\bigg)\bigg(\dfrac{1}{3}\bigg)\Bigg]=\dfrac{13}{6}}$

${\Rightarrow \sf 6-\Bigg[5-\dfrac{7}{6}\Bigg]=\dfrac{13}{6}}$

${\Rightarrow \sf 6-\Bigg[\dfrac{30-7}{6}\Bigg]=\dfrac{13}{6}}$

${\Rightarrow \sf 6-\Bigg[\dfrac{23}{6}\Bigg]=\dfrac{13}{6}}$

${\Rightarrow \sf \dfrac{36-23}{6}=\dfrac{13}{6}}$

${\Rightarrow \sf \dfrac{13}{6}=\dfrac{13}{6}}$

$\sf Hence\:Verified$

Required Answer :

$\boxed{\sf \dfrac{2}{3}(4x-1)-\bigg[2x-\dfrac{1+x}{3}\bigg]=\dfrac{1}{3}x+\dfrac{4}{3}=\dfrac{5}{2}}$