Question

PLs answer fast i need this questions answer

Answer 1

Answer:

$x=6$

Step-by-step explanation:

Given equation:

$x-\left(2x-\dfrac{3x-4}{7}\right)=\dfrac{4x-27}{3}-3$

Expand the left side:

$\implies x-2x+\dfrac{3x-4}{7}=\dfrac{4x-27}{3}-3$

Let all terms on the left side have the same denominator of 7:

$\implies \dfrac{7x}{7}-\dfrac{14x}{7}+\dfrac{3x-4}{7}=\dfrac{4x-27}{3}-3$

Join the terms on the left side:

$\implies \dfrac{7x-14x+3x-4}{7}=\dfrac{4x-27}{3}-3$

$\implies \dfrac{-4x-4}{7}=\dfrac{4x-27}{3}-3$

Let all the terms on the right side have the same denominator of 3:

$\implies \dfrac{-4x-4}{7}=\dfrac{4x-27}{3}-\dfrac{9}{3}$

Join the terms on the right side:

$\implies \dfrac{-4x-4}{7}=\dfrac{4x-27-9}{3}$

$\implies \dfrac{-4x-4}{7}=\dfrac{4x-36}{3}$

Cross multiply:

$\implies 3(-4x-4)=7(4x-36)$

Expand:

$\implies -12x-12=28x-252$

Add 12x to both sides:

$\implies -12=40x-252$

Add 252 to both sides:

$\implies 240=40x$

Divide both sides by 40:

$\implies 6=x$

$\implies x=6$