Solve (|x – 1| – 3) (|x + 2| – 5) < 0. Then -Maths 9th

Solve (|x – 1| – 3) (|x + 2| – 5) < 0. Then -Maths 9th
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Case 1. |x-1|-3>0 |x+2|–5<0 we need to find x-s that satisfy both inequalities. Case 2. |x-1|-3<0 |x+2|–5>0 we need to find x-s that satisfy both inequalities. After that we need to unite solutions of case 1 and case 2. In case 1 we have |x-1|>3, |x+2|<5 or x-1<-3, x-1>3 -5<x+2<5 or x<-2, x>4 -7<x<2 Draw both solutions on the axis. The common part is  -7<x<-2. Case 2 is treated similarly. The common solution is  3<x<4. The answer it the union of two intervals  -7<x<-2 and  3<x<4.
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