Question

Solve $x-\frac{2x+8}{3}=\frac{1}{4}(x-\frac{2-x}{6})-3$

Answer 1

Solve $x-\frac{2x+8}{3}=\frac{1}{4}(x-\frac{2-x}{6})-3$

$\therefore x-\frac{2x+8}{3}=\frac{1}{4}(x-\frac{2-x}{6})-3$

$\therefore \frac{3x-2x-8}{3}=\frac{x}{4}-\frac{2-x}{24}-3$

$\therefore \frac{x-8}{3}=\frac{6x-2+x-72}{24}$

$\therefore \frac{24}{3}(x-8)=7x-74$

$\therefore 8(x-8)=7x-74$

$\therefore 8x-64=7x-74$

$\therefore 8x-7x=-74+64$

$\therefore x=-10$