Решите дробно-рациональное уравнение

1) 2/x-3=5/x+4

2) 3/x+2=8/3-x

1) (2 (х+4)) / (х-3) = (5 (х-3)) / (х+4)

(2 х+8) / ((х-3) (х-4)) - (5 х-15) / ((х-3) (х+4)) = 0

(2 х+8-5 х+15) / ((х-3) (х+4)) = 0

(-3 х+23) / ((х-3) (х+4)) = 0

ОДЗ: (х-3) (х+4) ≠0; х≠3; х≠ - 4

-3 х+23 = 0

-3 х = - 23

х = - 23 : (-3)

х = 7 2/3

2) (3 (3-х)) / ((х+2) (3-х)) = (8 (х+2)) / ((х+2) (3-х))

(9-3 х) / ((х+2) (3-х)) - (8 х+16) / ((х+2) (3-х)) = 0

(9-3 х-8 х-16) / ((х+2) (3-х)) = 0

(-11 х-7) / ((х+2) (3-х)) = 0

ОДЗ: (х+2) (3-х) ≠0; х≠ - 2; х≠3

-11 х-7 = 0

-11 х = 7

х = 7 : (-11)

х = - 7/11