Question

Решите неравенство :
6(х+12)>3(х-4)
4(у-2)<5(у-3)
(3у+2)-3(2у+3)>12

Answer 1

6(х+12)>3(х-4)
6x + 72 > 3x - 12
3x > -84
x<-28


4(у-2)<5(у-3)
4y - 8< 5y - 15
y > 7

(3у+2)-3(2у+3)>12
3y + 2 - 6y - 9 > 12
-3y > 19
y< - 6 1/3