Respuesta :

[tex]\bf -\cfrac{2}{5}+\cfrac{2}{3}p+\cfrac{1}{3}p=2p-\cfrac{16}{5}\implies \stackrel{\textit{LCD of 15}}{-\cfrac{2}{5}+\cfrac{2p}{3}+\cfrac{p}{3}}=\stackrel{\textit{LCD of 5}}{2p-\cfrac{16}{5}} \\\\\\ \cfrac{(3\cdot -2)+(5\cdot 2p)+(5\cdot p)}{15}=\cfrac{(5\cdot 2p)-16}{5} \\\\\\ \cfrac{-6+10p+5p}{15}=\cfrac{10p-16}{5}\impliedby \textit{now we cross-multiply} \\\\\\ -30+50p+25p=150p-240\implies -30+240=150p-75p \\\\\\ 210=75p\implies \cfrac{210}{75}=p\implies \cfrac{14}{5}=p[/tex]

Otras preguntas