Answer:
[tex]x=6[/tex]
Step-by-step explanation:
Given equation:
[tex]x-\left(2x-\dfrac{3x-4}{7}\right)=\dfrac{4x-27}{3}-3[/tex]
Expand the left side:
[tex]\implies x-2x+\dfrac{3x-4}{7}=\dfrac{4x-27}{3}-3[/tex]
Let all terms on the left side have the same denominator of 7:
[tex]\implies \dfrac{7x}{7}-\dfrac{14x}{7}+\dfrac{3x-4}{7}=\dfrac{4x-27}{3}-3[/tex]
Join the terms on the left side:
[tex]\implies \dfrac{7x-14x+3x-4}{7}=\dfrac{4x-27}{3}-3[/tex]
[tex]\implies \dfrac{-4x-4}{7}=\dfrac{4x-27}{3}-3[/tex]
Let all the terms on the right side have the same denominator of 3:
[tex]\implies \dfrac{-4x-4}{7}=\dfrac{4x-27}{3}-\dfrac{9}{3}[/tex]
Join the terms on the right side:
[tex]\implies \dfrac{-4x-4}{7}=\dfrac{4x-27-9}{3}[/tex]
[tex]\implies \dfrac{-4x-4}{7}=\dfrac{4x-36}{3}[/tex]
Cross multiply:
[tex]\implies 3(-4x-4)=7(4x-36)[/tex]
Expand:
[tex]\implies -12x-12=28x-252[/tex]
Add 12x to both sides:
[tex]\implies -12=40x-252[/tex]
Add 252 to both sides:
[tex]\implies 240=40x[/tex]
Divide both sides by 40:
[tex]\implies 6=x[/tex]
[tex]\implies x=6[/tex]
