Question

Ejercicio Resuelto 145 del Algebra de Baldor - Numero 2: Hallar el numero que aumentado en sus 6/5 equivale a su triplo disminuido en 14

Answer 1

$x = \ n\'umero\qquad aumentado = suma\qquad tripo=3x \\ \\ x+ \frac{6}{5}x=3x-14 \\ \\ x+ \frac{6}{5}x-3x= -14 \\ \\ \frac{5}{5}x+ \frac{6}{5}x- \frac{3*5}{5}x= -14 \\ \\ \frac{5+6-15}{5}x= -14 \\ \\ - \frac{4}{5}x= -14 \\ \\ x= -14: - \frac{4}{5}\to \boxed{x=+ \frac{35}{2} }$

$Verificamos \\ \\ x+ \frac{6}{5}x=3x-14 \\ \\ \frac{35}{2} + \frac{6}{5} *\frac{35}{2} =3 *\frac{35}{2}-14 \\ \\ \frac{35}{2} + \frac{210}{10} =\frac{105}{2}-14 \\ \\ \frac{35*5}{2*5} + \frac{210}{10} =\frac{105}{2}- \frac{14*2}{2} \\ \\ \frac{175+210}{10} =\frac{105-28}{2} \\ \\ \frac{385}{10} =\frac{77}{2} \qquad \to\quad \frac{77}{2} =\frac{77}{2} \ \checkmark$

Espero que te sirva, salu2!!!!