Lenguaje coloquial a Lenguaje simbólico
La tercera parte de los seis quintos de un numero
[tex]\bf \dfrac{1}{3}*\dfrac{6}{5}x[/tex]
aumentados en cuatro quintos
[tex]\bf + \dfrac{4}{5}=[/tex]
los cinco medias partes de la diferencia entre cuatro tercios y ocho tercios
[tex]\bf \dfrac{5}{2}\left(\dfrac{4}{3} -\dfrac{8}{3} \right)[/tex]
Entonces
[tex]\bf \dfrac{1}{3}*\dfrac{6}{5}x\bf + \dfrac{4}{5}=\bf \dfrac{5}{2}\left(\dfrac{4}{3} -\dfrac{8}{3} \right)\qquad Resolvemos \\ \\ \\ \dfrac{1*6}{3*5}x\bf + \dfrac{4}{5}=\bf \dfrac{5}{2}\left(\dfrac{4-8}{3} \right)\\ \\ \\ \dfrac{6}{15}x\bf + \dfrac{4}{5}=\bf \dfrac{5}{2}\left(-\dfrac{4}{3} \right)\\ \\ \\ \dfrac{6}{15}x\bf + \dfrac{4}{5}=\bf -\dfrac{20}{6}\qquad Despejamos\\ \\ \\ \dfrac{6}{15}x = -\dfrac{20}{6}- \dfrac{4}{5}\\ \\ \\ \dfrac{6}{15}x = -\dfrac{20*5}{6*5}- \dfrac{4*6}{5*6}[/tex]
[tex]\bf \dfrac{6}{15}x = -\dfrac{100}{30}- \dfrac{24}{30}\\ \\ \\ \dfrac{6}{15}x = \dfrac{-100-24}{30}\\ \\ \\ \dfrac{6}{15}x = \dfrac{-124}{30}\\ \\ \\ x = \dfrac{-124}{30}: \dfrac{6}{15}\qquad Simplificamos \\ \\ \\ x = \dfrac{-31}{1}: \dfrac{3}{1}\\ \\ \\ x =- \dfrac{31}{3}[/tex]
Espero que te sirva, salu2!!!!!
