Реши уравнение используя основное свойство пропорции: 5+3x/12=4x-3/18; 0,9/7+5y=0,2/y-4...

1) $\frac{5+3x}{12} = \frac{4x-3}{18}$
$18(5+3x)=12(4x-3)$
$90+54x=48x-36$
$54x-48x=-36-90$
$6x=-126$
$x=-21$
2) $\frac{0.9}{7+5y} = \frac{0.2}{y-4}$
   $\frac{ \frac{9}{10} }{7+5y} = \frac{ \frac{2}{10}}{y-4}$
   $\frac{9}{10(7+5y)} = \frac{2}{10(y-4)}$
   $\frac{9}{10(7+5y)} = \frac{1}{5(y-4)}$
   $\frac{9}{10(7+5y)} = \frac{1}{5(y-4)}$ y≠-1.4, y≠4
   $45(y-4)=10(7+5y)$
   $45y-180=70+50y$
   $45y-50y=70+180$
   $-5y=250$
$y=-50$