Question

sa se rezolve in N ecuatiile:
a)(287x-2009):2009=2009
b)[(3x+5)*5-20]:10=8

varog complet

Answer 1

$a).(287x-2009):2009=2009 \\ 287x-2009=2009 \cdot 2009 \\ 287x-2009=4036081 \\ 287x=4036081+2009 \\ 287x=4038090 \\ x=4038090:287 \\ x=14070$
$b).[(3x+5)\cdot 5-20]:10=8 \\ (15x+25-20):10=8 \\ 15x-5=8 \cdot 10 \\ 15x+5=80 \\ 15x=80-5 \\ 15x=75 \\ x=75:15 \\ x=5$