Solving L.H.S of the equation 2x−35<53x+10≤54x+11; x∈R we get,
⇒2x−35<53x+10⇒36x−5<53x+50Multiplying both side by 15, we get:⇒15(36x−5)<15(53x+50)⇒5(6x−5)<3(3x+50)⇒30x−25<9x+150⇒30x−9x<25+150⇒21x<175⇒x<21175⇒x<325 ……..(1)
Solving R.H.S of the equation 2x−35<53x+10≤54x+11; x∈R we get,
⇒53x+10≤54x+11⇒53x+50≤54x+55Multiplying both side by 5, we get:⇒5(53x+50)≤5(54x+55)⇒3x+50≤4x+55⇒50−55≤4x−3x⇒−5≤x⇒x≥−5 ……..(2)
From equation (1) and (2), we get :
-5 ≤ x < 325.
Hence, solution set equals to -5 ≤ x < 325, x ∈ R.