Question

(0,6у-1)-0,2(3у+1)<5у-4

Answer 1

$(0,6y-1)-0,2(3y+1)\ \textless \ 5y-4 \\ \\ 0.6y-1-0,6y-0,2\ \textless \ 5y-4 \\ \\ -1,2\ \textless \ 5y-4 \\ \\ -5y\ \textless \ -4+1,2 \\ \\ -5y\ \textless \ -2,8 \\ \\ y\ \textgreater \ 0,56$

x∈$(0,56; +\infty});$