Question

55+\frac{x}{4}=\frac{1}{3}(2-\frac{x}{2} )-\frac{2}{3} +\frac{1}{4} (10-\frac{5x}{3} )

Answer 1

$55+\frac{x}{4}=\frac{1}{3}(2-\frac{x}{2} )-\frac{2}{3} +\frac{1}{4} (10-\frac{5x}{3} )$
 
(se resta de ambos lados 55)

$-55 +55+\frac{x}{4}=\frac{1}{3}(2-\frac{x}{2} )-\frac{2}{3} +\frac{1}{4} (10-\frac{5x}{3} ) -55$

$\frac{x}{4}=-\frac{167}{3} +\frac{-7x+38}{12}$

Aplicar el Mínimo Común Múltiplo 12 

$\frac{x}{4}*12=-\frac{167}{3}*12 +\frac{-7x+38}{12}*12$

$\frac{12x}{4}=-\frac{167*12}{3} +\frac{-7x+38*12}{12}$

$3x=-668 -7x +38$

$3x+7x = -630$

$10x = -630$

Dividir 10 en ambos lados.

$\frac{10x}{10} = \frac{-630}{10}$

x = -63