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5(x+2)+(x-3)<3(x-1)+4x

Răspuns :

5x+10+x-3<3x-3+4x
6x-7x<-3+3-10
-x<-10
x>10
solutie :e apartine intervalului (10 infinit)
5x+10+x-3 < 3x-3+4x
6x-7 < 7x-3
-x-4 < 0
-x < 4
x < -4