Question

решите неравенство 
5(4y+3) - 7 (3y-4)≤ 10 
2(z-2)/9≤ 3+z/7
12-y< 5(y-1)/6

Answer 1

20у+15-21у+28<=10
-y+43<=10
-y<=10-43
-y<=-33
y=>33
Ответ (-бесконечности; 33]
7*2(z-2)<=9(3+Z)
14z-28<=27+9z
14z-9z<=27+28
5z<=55
z<=55/5
z<=11
[11;+ бесконечности)
6*(12-y)<5(y-1)
72-6y<5y-5
-6y-5y<-5-72
-11y<-77
y>-77/-11
y>7
(-бесконечности,7)